Flag-Transitive and Almost Simple Orbits in Finite Projective Planes

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◭ ◮ Flag-transitive and almost simple orbits in page 1 / 37 ﬁnite projective planes go back Alessandro Montinaro full screen

Abstract close Let Π be a projective plane of order n and let G be a collineation group quit of Π with a point-orbit O of length v. We investigate the triple (Π, O, G) when O has the structure of a non trivial 2-(v,k, 1) design, G induces a ﬂag- transitive and almost simple automorphism group on O and n ≤ P(O) = b + v + r + k.

Keywords: projective plane, collineation group, orbit MSC 2000: 51E15, 20B25

1. Introduction and statement of the result

A classical subject in finite geometry is the investigation of a finite projective plane Π of order n admitting a collineation group G which acts doubly transitively (flag-transitively) on a point-subset of size v of Π. It dates back to 1967 O and it is due to Cofman [9]. It is easily seen that either (i) the structure of a non-trivial 2-(v, k, 1) design (i.e. k 3 and at least two ≥ distinct blocks) is induced on , or O (ii) is an arc, or O (iii) is a contained in a line. O The current paper entirely focuses on the situation in which is a non-trivial O 2-(v, k, 1) design. Several papers deal with the case n v. Conclusive results ≤ where obtained by Luneburg¨ [35], in 1966, when is a Ree unital, and later on O by Kantor [28], by Biliotti and Korchmaros´ [5, 6], when is a Hermitian unital. O ACADEMIA Characterization results were also provided by Ostrom-Wagner [43] in 1959 and PRESS by Luneburg¨ [37] in 1976 when is a projective subplane of Π. O I I G

A fundamental contribution to the problem is the classification of the 2-tran- ◭◭ ◮◮ sitive non-trivial 2-(v, k, 1) designs due to Kantor [30] in 1985, and some years later, the classification of the flag-transitive non-trivial 2-(v, k, 1) designs due to ◭ ◮ Buekenhout, Delandtsheer, Doyen, Kleidman, Liebeck and Saxl [2, 3, 34, 47]. Recently, Biliotti and Francot [4] investigated the general case when n v and page 2 / 37 ≤ G acts doubly transitively on , determining all possible collineation groups. O go back The problem of classifying the triple (Π, , G) in the case where is a non- O O trivial 2-(v, k, 1) design for n>v, but v close to n, has also been investigated. full screen Indeed, Dempwolff [11] in 1985 dealt with the case = PG(2, q) and n = q3. O ∼ Such a case has been recently generalized by Montinaro [39] to n q3. More- ≤ close over, the author [40] shows that cannot be a Ree unital of order q when O n q4, generalizing the result of Luneburg¨ [37]. Finally, in 2005, Biliotti and ≤ quit Montinaro [8] showed that no solutions to the above problem involving non- trivial 2-(v, k, 1) designs arise for n = v + 3. In this paper we investigate the triples ( , Π, G) under the assumption that O is a non-trivial 2-(v, k, 1) design, G is a collineation group of Π inducing an O almost simple and flag-transitive automorphism group on and n ( ), O ≤ O where ( ) = b + v + r + k. Clearly b denotes the number of blocks ofP and O O r the numberP of blocks of incident with any given point of . In particular, O O our result is contained in Theorem 1.1.

Let be any set of points in Π. A line l of Π is called an external line, D or a tangent or a secant to , according to l = 0, 1 or k, with k > 1, D |D ∩ | respectively. We may regard as an incidence structure, where the blocks are D the secants of . Any incidence structure represented in this way is said to be D embedded in Π. In particular, any orbit of a collineation group G of Π may be O regarded as an embedded structure. A flag is an incident point-line pair of Π, so is said to be flag-transitive, if G is transitive on the flags of . Finally a O O flag-transitive orbit is said to be almost simple, if the group induced by G on O is almost simple (i.e. G has a non-abelian simple normal subgroup L such that L E G E Aut(L)).

Theorem 1.1. Let Π be a projective plane of order n and let G be a collineation group of Π with a point-orbit of length v. If has the structure of a non- O O trivial 2-(v, k, 1) design, the group G induces a ﬂag-transitive and almost simple automorphism group on and n ( ), then one of the following holds. O ≤ O (1) G acts faithfully on and one ofP the following occurs: O (a) v = n2 + n + 1, = Π = PG(2, n) and PSL(3, n) G; ACADEMIA O ∼ ≤ PRESS (b) v = n + √n + 1, = PG(2, √n), Π is a Desarguesian plane or a gener- O ∼ alized Hughes plane, and PSL(3, √n) G; ≤ I I G

(c) v = 7, = PG(2, 2), Π = PG(2, 8) and PSL(3, 2) G; ◭◭ ◮◮ O ∼ ∼ ≤ (d) v = 7, = PG(2, 2), Π = PG(2, 16) and PSL(3, 2) G; O ∼ ∼ ≤ ◭ ◮ (e) v = 13, n = 33, = PG(2, 3) and PSL(3, 3) G; O ∼ ≤ (f) v = √n3 + 1, = (√n), Π = PG(2, n) and PSU(3, √n) G; page 3 / 37 O ∼ H ∼ ≤ (g) v = √4 n3 +1, = (√4 n) is contained in a plane Π = PG(2, √n) which O ∼ H 0 ∼ go back is left invariant by G, and PSU(3, √4 n) G; ≤ (h) v = n (n 1), n = 2r, = (n), Π = PG(2, n) and PSL(2, n) G. Here full screen 2 − O ∼ W ∼ ≤ (n) denotes the Witt space associated with PSL(2, n). Furthermore, the W projective extension of is embedded in Π and the set of external lines close O C to is a line-hyperoval extending a line-conic of Π; O √n 2r quit (i) v = (√n 1), n = 2 , = (√n) and PSL(2, √n) G; 2 − O ∼ W ≤ √n 1 r 2 (j) v = − √n, n = (2 + 1) , = (√n 1) and PSL(2, √n 1) G. 2 O ∼ W − − ≤ (2) G does not act faithfully on and one of the following occurs: O (a) v = 75, = PG(2, 7), Π is the generalized Hughes plane over the excep- O ∼ tional nearﬁeld of order 72 and SL(3, 7) G; ≤ (b) v = √4 n3 + 1, √4 n3 2 mod 3, = (√4 n) is contained in a Desargue- ≡ O ∼ H sian Baer subplane and SU(3, √4 n) G. ≤ Furthermore, in each case, except in (1j) and possibly in (1i), the involutions in G are perspectivities of Π.

Cases (1a)–(1d), (1f), (1h) and (2a) really occur. Case (1e) occurs in the Desarguesian plane and in the Hering-Figueroa plane of order 33 (see [14] and [19]). Furthermore, case (1g) occurs in the Desarguesian or Hughes planes, and case (1i) occurs in the Desarguesian plane. Finally, cases (1j) and (2b) are open. Some restrictions about the possible existence of an example for case (1j) are provided in Corollary 3.10.

2. Preliminaries and background

We shall use standard notation. For what concerns finite groups the reader is referred to [15] and [25]. The necessary background about finite projective planes and 2-(v, k, 1) designs may be found in [24] and in [49], respectively. Let Π=( , ) be a finite projective plane of order n. If H is a collineation P L ACADEMIA group of Π and P (l ), we denote by H(P ) (by H(l)) the subgroup of H PRESS ∈P ∈ L consisting of perspectivities with center P (axis l). Also, H(P, l) = H(P ) H(l). ∩ Furthermore, we denote by H(P,P ) (by H(l, l)) the subgroup of H consisting I I G

of elations with center P (axis l). A collineation group H of Π is said to be ◭◭ ◮◮ irreducible on Π, if H does not ﬁx any point, line, or triangle of Π. An irreducible collineation group H is said to be strongly irreducible on Π, if H does not ﬁx any ◭ ◮ proper subplane of Π.

page 4 / 37 Let X be a collineation group of Π and recall that Fix(X) denotes the subset of Π consisting of the points and of the lines of Π which are fixed by X. If X go back is planar, i.e. Fix(X) is a subplane of Π, we denote by o(Fix(X)) the order of plane Fix(X). full screen Let be a H-orbit of points in Π on which the structure of a non-trivial N 2-(v, k, 1) design can be induced. Suppose that admits a parallelism, that is N close the blocks of are partitioned into parallel classes such that each class is a N partition of the points of . If for each pencil Φ of parallel blocks, the blocks quit N of Φ pass through a common point PΦ of Π, and all PΦ’s lie in the same line l N of Π, then, following [13] we say that the projective extension of is embedded N in Π. Before starting our investigation we recall some combinatorial and group- theoretical results which are useful hereafter. The following theorem deals with the general structure of G. Theorem 2.1 (Buekenhout et al. [2]). Suppose that G is a flag-transitive group of automorphisms of a non-trivial design . Then either D (I) G is almost simple: that is, G has a non-abelian simple normal subgroup N such that N E G E Aut(N), or (II) G is of affine type: that is, the set of points of carries the structure of an D affine space AG(t, p) which is invariant under G, and G contains the whole translation group T of AG(t, p), (so T E G AGL(t, p)). ≤ The non-trivial 2-(v, k, 1) designs admitting a flag-transitive automorphism D group G are classified in [47] when G is of type (I) and in [34] when G is of type (II), respectively. In our analysis we focus entirely on case (I) and we have: Theorem 2.2 (Saxl [47]). Let be a design and suppose that G is a flag-transitive D almost simple group of automorphisms of . Then one of the following occurs: D PG PSL G PΓL (i) ∼= (d, q), d 2, and either (d + 1) (d + 1, q) or D G≥ A ≤ ≤ (d, q)=(3, 2) and ∼= 7; (ii) = (q) is a Hermitian unital and PSU(3, q2) G PΓU(3, q); D ∼ H ≤ ≤ h 2G G (iii) ∼= (q), q = 3 , h odd, h 1, is a Ree unital associated to 2 (q) D2 R ≥ ≤ ≤ ACADEMIA G2 (q) .Zh; PRESS (iv) = (q), q = 2h 8, is a Witt space associated to PSL(2, q) G D ∼ W ≥ ≤ ≤ PΓL(2, q). I I G

We shall make extensive use the following deﬁnition and lemma, which are ◭◭ ◮◮ independent of the context of non-trivial 2-(v, k, 1) designs embedded in projective planes. ◭ ◮ Deﬁnition 2.3. Let be a non-trivial 2-(v, k, 1) design. Then page 5 / 37 D ( ) = b + v + r + k, D go back P v 1 where r = k−1 is the number of blocks which are incident with any given point − full screen v(v 1) of and b = k(k−1) is the total number of blocks of . D − D close Lemma 2.4. The following relations hold: q2d+1 + qd+3 + qd+2 2qd+1 2qd + q4 4q2 + 4 (1) PG(d, q) = − − − ; quit (q2 1) (q 1) P − − (2) (q) = q4 + 2q2 + q + 2 ; H P (3) (q) = q4 + 2q2 + q + 2, where q = 3h and h is odd, h 1 ; R ≥ P q (4) (q) = (3q + 2), where q = 2h 8 . W 2 ≥ P Proof. Omitted.

The following theorem, which relies on the results of Ostrom-Wagner [43], Luneburg¨ [37] and Dempwolff [11], deals with the case = PG(2, q) and O ∼ n q3. ≤ Theorem 2.5 ([39]). Let Π be a ﬁnite projective plane of order n and let G be a collineation group of Π inducing a group containing PSL(3, q) on a subplane Π0 of order q. If n q3, then one of the following occurs: ≤ (1) Π = PG(2, q), PSL(3, q) G and one of the following occurs: 0 ∼ ≤ (a) n = q, and Π = Π0; 2 (b) n = q , Π is a Desarguesian plane or a generalized Hughes plane and Π0 is a Baer subplane of Π; (c) n = q3. PG (2) Π0 ∼= (2, 7), Π is the generalized Hughes plane over the exceptional nearﬁeld of order 72 and SL(3, 7) G. ≤ Based on a similar idea, the following theorem generalizes a result of Luneburg¨ ACADEMIA PRESS [35]. I I G

Theorem 2.6 ([40]). If (q), q = 3h, is embedded in a ﬁnite projective plane ◭◭ ◮◮ R Π of order n, with n q4, in such a way that 2G (q) is induced on (q) by a ≤ 2 R collineation group G of Π. Then h = 1, G acts faithfully on (3) as PΓL(2, 8) and ◭ ◮ R one of the following occurs: page 6 / 37 (1) Π = PG(2, 8), G leaves a line oval of Π invariant and consists of the ∼ C O external points of ; go back C (2) n = 26.

full screen We conclude the preliminaries with the following numerical lemma that will be useful in dealing with the case when is an Hermitian unital of order 22h, close O h > 1.

quit Lemma 2.7. Let f, h, t and λ1 be positive integers. Then the Diophantine equation

22h 1 f − =1+2tλ (2h+1 + 2tλ ) (1) 3 1 1

h t has no solutions for h > 1, t = (h + 1)/2 and λ 2 − . ⌊ ⌋ 1 ≤ Proof. Set a = f/3. Then (1) becomes

a(22h 1) = 1 + 2tλ (2h+1 + 2tλ ) . (2) − 1 1 h t t h+1 t 2h As λ 2 − , then 1 + 2 λ (2 + 2 λ ) 1 + 2 3. Then a 4 by (2). 1 ≤ 1 1 ≤ ≤ Assume that (f, 3) = 3. Then a is an integer. Since the second part of (2) is odd and since a 4, then either a = 1 or 3. Assume that a = 1. Then 22h = t h+1 ≤t 2 t h+1 t 2 + 2 λ1(2 + 2 λ1) by (2). A contradiction, since 2 ∤ 2 + 2 λ1(2 + 2 λ1) as t 1 while h > 1. So a = 3. Then 22h3=4+2tλ (2h+1 + 2tλ ) again by (2). If ≥ 1 1 t 2, then 23 ∤ 4 + 2tλ (2h+1 + 2tλ ). This forces h = 1. A contradiction by our ≥ 1 1 assumption. Thus t = 1 and hence h = 2 and λ1 = 1, 2, since t = (h + 1)/2 h t ⌊ ⌋ and 1 λ 2 − . So, substituting t = 1, h = 2 and λ = 1, 2 in (2), we obtain ≤ 1 ≤ 1 a contradiction since a must be a positive integer as (f, 3) = 3. Assume that (f, 3) = 1. Then f 12 as a = f/3 and a 4. Furthermore t h+1 t ≤ ≤ f is odd, since 1 + 2 λ1(2 + 2 λ1) in (1) is so. All these informations yield f 1, 5, 7, 11 . Now, by managing (1), we obtain ∈ { } 2h t h+1 t 2 f = (3+ f) + 2 3λ1(2 + 2 λ1) . (3)

2 t h+1 t If f 7, 11 , then 2 ∤ 2 + 2 λ1(2 + 2 λ1) as t 1. A contradiction, since ∈ { } ≥2h t h+1 t ACADEMIA h > 1. Hence f 1, 5 . If f = 1, then (3) becomes 2 = 4+2 λ1(2 +2 λ1). PRESS ∈ { } If t 2, then 23 ∤ 4 + 2tλ (2h+1 + 2tλ ). A contradiction, since h > 1 by our ≥ 1 1 assumptions. So t = 1 and hence h = 2 and λ = 1, 2 as h > 1, t = (h + 1)/2 1 ⌊ ⌋ I I G

h t and 1 λ1 2 − by our assumptions. Now, substituting these values in (3), ◭◭ ◮◮ ≤ ≤ we obtain a contradiction since f = 1. Thus f = 5. Then (3) becomes 22h5 = 23 + 2t3λ (2h+1 + 2tλ ). If t 2, then 24 4 + 2tλ (2h+1 + 2tλ ). This yields ◭ ◮ 1 1 ∤ 1 1 4 2h 2h 3 ≥ t h+1 t that 2 ∤ 2 5 as 2 5 = 2 + 2 3λ1(2 + 2 λ1). A contradiction. So t = 1, h t h = 2 and λ = 1, 2, as t = (h + 1)/2 and 1 λ 2 − . By substituting page 7 / 37 1 ⌊ ⌋ ≤ 1 ≤ these values in (3), we obtain again a contradiction as f = 5. This completes go back the proof.

Now, let be a ﬁnite projective plane of order and let be a collineation full screen Π n G group of Π with a ﬂag-transitive and almost simple point-orbit . Let N be O the kernel of the action of G on . As a consequence of the our assumptions, close O the pair ( , G/N) is listed in Theorem 2.2. So, in order to classify the triples O (Π, , G) when n ( ), we treat the cases N = 1 and N = 1 separately. quit O ≤ O h i 6 h i P 3. The faithful action

Throughout this section we assume that N = 1 . Hence G acts faithfully on . h i O Lemma 3.1. Let Π be a projective plane of order n and let G be a collineation group of Π with a point-orbit = PG(d, q), on which G acts faithfully and induces O ∼ a ﬂag-transitive almost simple automorphism group. If n ( ) then (d, q) = ≤ O 6 (3, 2). P

PG A PSL Proof. Assume ∼= (3, 2). Then either G ∼= 7 or (4, 2) G by Theo- AO PSL A ≤ rem 2.2(i). As 7 < (4, 2), we may assume G ∼= 7. Note that n 60 as (PG(3, 2)) = 60 by Lemma 2.4(1). On the other hand, ≤ n > 15 by [4]. ThereforeP 16 n 60. Let P . Then GP = PSL(2, 7) by ≤ ≤ ∈ O ∼ [1]. Now, let σ be an arbitrary involution in GP . Then σ ﬁxes exactly 3 of the 5 secants to through P (see [42]). Thus σ is a Baer involution of Π and hence O n is a square, namely n = 16, 25 or 49. Assume n = 16. Since G is transitive on P and since P = 14, P O − { } |O − { }| we have a contradiction by [12, 4.6(a), (b), 4.8(i)].

Assume n = 25. Let ϕ be any element of order 7 in GP . Then ϕ ﬁxes at least 5 lines of [P ] and 2 lines of Π [P ], as n + 1 5 mod 7 and n2 2 − ≡ ≡ mod 7. Thus ϕ is planar with o(Fix(ϕ)) = 4+7θ, where θ 0. Actually, ≥ θ = 0 by [24, Theorem 3.7], since n = 25. Hence o(Fix(ϕ)) = 4. Note that N ( ϕ ) = ϕ, ψ where o(ψ) = 3. Also N ( ϕ ) is the unique maximal GP h i h i GP h i ACADEMIA subgroup of GP = PSL(2, 7) containing ϕ. Hence for each line u Fix(ϕ) [P ], PRESS ∼ ∈ ∩ either G = ϕ or G = ϕ, ψ or G = G . Assume that G = P,u h i P,u h i P,u P P,m ϕ for some line m Fix(ϕ) [P ]. Then mGP = 24. Furthermore, we h i ∈ ∩

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have sGP = 7 for any line s [P ] . Hence mGP + sGP 31. On ◭◭ ◮◮ ∈ ∩ O ≥ the other hand, mGP + sGP 26 as n + 1 = 26 , mG P sGP [P ] and

GP GP ≤ ∪ ⊆ m s = . A contradiction. Thus either GP,u = ϕ, ψ or GP,u = GP for ◭ ◮ ∩ ∅ h i each line u Fix(ϕ) [P ]. Thus Fix(ϕ) [P ] = Fix( ϕ, ψ ) [P ] and hence ∈ ∩ ∩ h i ∩ Fix( ϕ, ψ ) [P ] = 5. Assume that Fix( ϕ, ψ ) [P ] Fix(G ) [P ] 3. page 8 / 37 | h i ∩ | | h i ∩ − P ∩ | ≥ Let u , i = 1, 2, 3, be some of the lines of [P ] ﬁxed by the group ϕ, ψ but i h i GP go back not by the group GP . Then ui = 8 for each i = 1, 2, 3. Furthermore, as

ϕ, ψ is maximal in G , the line u is the unique line in uGP which is ﬁxed h i P i i full screen by ϕ, ψ for each i = 1, 2, 3. Thus uGP uGP = ∅ for each h, j = 1, 2, 3 h i h ∩ j with h = j. Then there are at least three G -orbits on [P ] each of length 6 P close 8. So these G -orbits involve 24 of the lines of [P ]. Then n + 1 24 + 7, P ≥ since sGP = 7 for any s [P ] . A contradiction, since n = 25. Thus quit ∈ ∩ O Fix( ϕ, ψ ) [P ] Fix(G ) [P ] 2 and hence Fix(G ) [P ] 3. Now, | h i ∩ − P ∩ | ≤ | P ∩ | ≥ we may repeat the above argument with l in the role of [P ] for each line l ∈ Fix(G ) [P ]. This yields Fix(G ) l 3 for each line l Fix(G ) [P ]. P ∩ | P ∩ | ≥ ∈ P ∩ Then GP is planar and Fix(GP ) is a proper subplane of Fix(σ) since ϕ and σ ﬁx exactly 0 and 3 lines in [P ] , respectively. That contradicts [24, Theorem ∩O 3.7] since Fix(σ) has order 5. PSL Finally, assume that n = 49. Then GP ∼= (2, 7) contains a unique conjugate class of involutions which are homologies of Π by [22]. A contradiction, since σ G and σ is a Baer involution of Π. ∈ P Lemma 3.2. Under the same assumptions as in Lemma 3.1, we have d = 2.

Proof. Assume d 3. Then PSL(d + 1, q) G by Theorem 2.2(i) and by ≥ ≤ Lemma 3.1, since G is faithful, flag-transitive and almost simple on . Since O all the assumptions are satisfied by the subgroup of G which is isomorphic to PSL PSL (d + 1, q), we may assume G ∼= (d + 1, q). Let T be a subgroup of G consisting of the projective transvections of PSL(d + 1, q) fixing the same hyper- plane of pointwise, as d 3. That is Fix(T ) = PG(d 1, q). Hence T is O ≥ ∩O ∼ − planar. Set Fix (T ) = Fix(T ) . Clearly, Fix (T ) Ψ Fix(T ). Assume O ∩O O ⊆ ⊆ d > 3. Then Fix (T ) Ψ, since Fix (T ) = PG(d 1, q). Let W be any given O O d−∼1 ⊂ q 1 − point of Fix (T ). Then there are exactly − lines of Fix (T ) PG(d 1, q) q 1 ∼= O − O − − qd 1 1 through W by [49, Theorem 1.4.10]. Thus, there are at least q 1− + 1 lines − − qd 1 1 of Ψ through W , since Fix (T ) Ψ. Therefore, o(Ψ) q 1− . If Ψ = Fix(T ), O ⊂ ≥ PSL− let us consider the subgroup J of NG(T ) such that J ∼= (d, q). Then J acts doubly transitively on Fix (T ) = PG(d 1, q) and acts faithfully on Fix(T ). If d O ∼ q 1 − ACADEMIA o(Fix(T )) q −1 , we obtain a contradiction by [4, Theorem 3.13 and Propo- PRESS ≤ − qd 1 sition 3.14]. Hence, we may assume that o(Fix(T )) > q −1 in this case. If − I I G

− 2 qd 1 1 is a proper subplane of , then − by [24, The- ◭◭ ◮◮ Ψ Fix(T ) o(Fix(T )) q 1 ≥ − qd 1 orem 3.7]. In particular, this implies o(Fix(T )) q −1 as d > 3. Therefore, ◭ ◮ ≥ − qd 1 o(Fix(T )) q −1 in any case, and ≥ − page 9 / 37 2 qd 1 − n ( ) (4) go back q 1 ≤ ≤ O − P by [24, Theorem 3.7]. Hence full screen (qd 1)2 q2d+1 + qd+3 + qd+2 2qd+1 2qd + q4 4q2 + 4 close − n − − − (5) (q 1)2 ≤ ≤ (q2 1) (q 1) − − − quit by Lemma 2.4(1). Multiplying each term of (5) by (q2 1)(q 1) we obtain − − q2d qd+3 qd+2 q4 + 4q2 + q 3 0 . (6) − − − − ≤ That is d+2 d 3 2 4 q q(q − 1) 1 + (4q + q 3) q 0 , (7) − − − − ≤ which is impossible as d > 3. Now, we rule out the case d = 3 in ﬁve steps.

(I) q is odd.

Assume that q is even. Then τ is a Baer involution of Π. Note that Fix(τ) ∩ = PG(2, q) as d = 3. Let us consider the subgroup R of C (τ) such that R = O ∼ G ∼ PSL(3, q). Then R acts on Fix(τ). Moreover, R acts in its natural 2-transitive permutation representation on Fix(τ) = PG(2, q). Lemma 2.4(1) with d = 3 ∩O ∼ yields ( ) = q4 + 2q3 + 4q2 + 4q + 4, and then, by using (4), that q4 < n < O (q + 2)P4. This yields √n < q3 for q > 2. Furthermore, √n < q3 for q = 2 by a direct substitution in (4). Hence, we may apply Theorem 2.5, with Fix(τ) in the role of Π, to assert that either √n = q or √n = q2. A contradiction in any case as n > q4. Thus q is odd.

(II) If G fixes a point of Π, then the involution σ in G represented by the matrix diag( I ,I ) is a Baer collineation of Π. − 2 2 Assume that G fixes a point Q on Π. Clearly Q Π . Then G acts on [Q]. ∈ −O Note that each line through Q intersecting is a tangent to , as G is primitive O O on . As q is odd by (I), let σ be the involution in G represented by the matrix O ACADEMIA diag( I ,I ). Then σ fixes pointwise two skew lines on and hence σ fixes PRESS − 2 2 O exactly 2(q + 1) tangents to through Q. Therefore, σ is a Baer collineation qO4 1 of Π, since there are exactly q −1 tangents to through Q. − O I I G

(III) If G fixes a line of Π, then it fixes a point on this line. ◭◭ ◮◮ Assume that G fixes a line l of Π. Clearly l is external to . We assume that ◭ ◮ O G does not fix points on l and this leads to a contradiction, as we are going to

page 10 / 37 see. Assume that distinct lines of intersect l in distinct points. Then there exists O go back X l such that XG = b, as G is transitive on the lines of . If l = XG, ∈ O then the commuting involutions σ and σ′ = diag(1, 1, 1, 1) are homologies − − full screen of Π, since they fix exactly two points on l. Indeed, the actions of σ and σ′ on XG and on the lines of PG(3, q) are the same, but they have no common fixed close points on l. A contradiction by [29, Lemma 3.1]. Then there exists Y l XG. 4 G G q∈ 1− Then Y > 1 by our assumption. Therefore Y v, where v = − is the ≥ q 1 quit minimal primitive permutation representation degree of G = PSL(4,− q). Note ∼ that l (XG Y G) r + k + 1, since Y G v, XG = b and n ( ). − ∪ ≤ 4 ≥ 3 ≤ O q 1 q 1 P Now observe that v > r + k + 1 being v = q − 1 , r = q −1 and k = q + 1. Thus, − − if l = XG Y G, the group G fixes l (XG Y G) pointwise, since any non- 6 ∪ −q4 1 ∪ trivial G-orbit has length at least v = q −1 . That contradicts our assumptions. − Hence l = XG Y G. Then either Y G = v or Y G > v. Assume the latter ∪ occurs. Then Y G qv by [33, Lemma 4.2 and Table II], for G > 1012 and ≥ | | by [31] for G 10 12. Therefore qv v + r + k + 1 as Y G l XG. Again a 4 3 | | ≤ q 1 q ≤1 ⊆ − G G contradiction, since v = q −1 , r = q −1 and k = q + 1. So l = X Y with − − ∪ Y G = v and XG = b. Then σ fixes exactly q + 3 points on l. Namely q + 1 points on Y G and 2 points on XG (this follows by the fact that the actions of

σ on Y G and on XG are the same as those on the points and on the lines of PG(3, q), respectively). So σ is a Baer collineation and hence n = (q + 2)2. A 4 q 1 2 2 contradiction, since n = b + v 1, with v = − and b = (q + 1)(q + q + 1) − q 1 as = PG(3, q). − O ∼ Hence there are two distinct secants to , say s and s , concurring at a O 1 2 point P . Then G , G G , as G ﬁxes l. Therefore G , G G . Then s1 s2 ≤ P h s1 s2 i ≤ P G = GP , since Gs1 and Gs2 are two distinct maximal parabolic subgroups of G (see [32] and [31]). That contradicts our assumptions. This completes the proof of (III).

(IV) The group G contains involutory perspectivities.

Let σ be the involution in G represented by the matrix diag( I ,I ). Assume − 2 2 that σ is a Baer collineation of Π. Now, consider the subgroup H of CG(σ) ACADEMIA consisting of the matrices diag(A, I ), where A PSL(2, q). Then H = PSL(2, q) PRESS 2 ∈ ∼ and H acts on Fix(σ). In particular there exists a secant u to such that σ and O H ﬁx the q + 1 points of u . ∩O I I G

If H contains Baer collineations of Fix(σ), then n must be a fourth power. ◭◭ ◮◮ Note that, by the first inequality of (5), we have n > q4. Then √4 n > q and hence √4 n q +1, since √4 n is an integer. Thus n (q +1)4. On the other hand, ◭ ◮ ≥ ≥ n ( ) = q4 + 2q3 + 4q2 + 4q + 4. So (q + 1)4 q4 + 2q3 + 4q2 + 4q + 4. ≤ O ≤ A contradiction. Thus each involution in is a perspectivity of , since page 11 / 37 P H Fix(σ) H contains a unique conjugate class of involutions. In particular each involution in H has axis u Fix(σ), since σ and H fix the q + 1 points of u by the above go back ∩ ∩O argument. Thus H = H(u Fix(σ)), since H is generated by its involutions. ∩ full screen Assume q > 3. Then H = H(C, u Fix(σ)) for some point C Fix(σ) ∩ ∈ by [24, Theorems 4.14 and 4.25], since H is non-abelian simple. Then either close H √n or H √n 1 according to whether C does or does not lie on | | | | | | − u Fix(σ) respectively. This yields √n q(q2 1)/(q 1, 2) in any case as ∩ ≥ − − quit H = q(q2 1)/(q 1, 2). Thus | | − − q(q2 1) 2 − q4 + 2q3 + 4q2 + 4q + 4 (8) (q 1, 2) ≤ − as n q4 + 2q3 + 4q2 + 4q + 4. Inequality (8) yields q 3, contrary to our ≤ ≤ assumption. Assume q 3. Actually q = 3 by Lemma 3.1 as d = 3. Then ( ) = 187 and ≤ O hence n 132 as n is a square and n ( ). On the other handP n > 34 by the ≤ ≤ O first inequality of (5), as q = 3. Thus √nP= 10, 11, 12 or 13, since n is a square. The case √n = 10 cannot occur by [24, Theorem 13.18]. Finally, if √n = 12, A then H ∼= 4. A contradiction by [27]. Hence either √n = 11 or √n = 13. Now, consider the group R = H L where H is defined as above (for q = 3) and L × consists of the matrices diag(I ,B), where B PSL(2, 3). Then R = A A 2 ∈ ∼ 4 × 4 and R C (σ). In particular R acts faithfully on F ix(σ). As √n = 11 or ≤ G 13, then R contains a subgroup R0 of homologies of Fix(σ) isomorphic to E16. Then R (D, a) for some point D Fix(σ) and some line a Fix(σ) by [29, 0 ∈ ∈ Lemma 3.1]. Thus 16 (√n 1). A contradiction, since √n = 11 or 13. Thus | − the assertion (III).

(V) The ﬁnal contradiction.

By (IV) the group G contains involutory perspectivities of Π. In particular the proof of (IV) implies that the involution σ in G represented by the matrix diag( I ,I ) is a perspectivity of Π. This yields that G does not ﬁx points or − 2 2 lines by (II) and (III) respectively. This yields in turn that G does not ﬁx tri- PSL angles, since G is simple. Therefore the group G ∼= (4, q) is an irreducible ACADEMIA collineation group of Π containing involutory perspectivities. A contradiction PRESS by [48].

Thus d = 2 and hence the assertion. I I G

Lemma 3.3. Let Π be a projective plane of order n and let G be a collineation ◭◭ ◮◮ group of Π with a point-orbit = PG(d, q), on which the group G acts faithfully O ∼ and induces a flag-transitive almost simple automorphism group. If n ( ) ◭ ◮ ≤ O then d = 2 one of the following occurs: P page 12 / 37 (1) n = q, Π = = PG(2, q) and PSL(3, q) G; O ∼ ≤ (2) n = q2, Π is either a Desarguesian plane or a generalized Hughes plane, go back = PG(2, q) is a Baer subplane of Π and PSL(3, q) G; O ∼ ≤ (3) n = 8, q = 2, Π = PG(2, 8), = PG(2, 2) and PSL(3, 2) G; full screen ∼ O ∼ ≤ PG PG PSL (4) n = 16, q = 2, Π ∼= (2, 16), ∼= (2, 2) and (3, 2) G; close O ≤ (5) n = 27, q = 3, = PG(2, 3) and PSL(3, 3) G. O ∼ ≤ quit Proof. Clearly = PG(2, q) by Lemma 3.2. Then n ( ) = 2q2 + 4q + 4 O ∼ ≤ O by Lemma 2.4(1). If q > 3 then we have 2q2 + 4q + 4 P q3. Hence if q > 3 ≤ or if q 2, 3 , n q3 we may apply Theorem 2.5 and obtain assertions (1), ∈ { } ≤ (2), (3) and (5) of our statement. We assume now q = 2 or 3; the inequalities q3 < n 2q2 + 4q + 4 yield 9 n 20 or 28 n 34, respectively. ≤ ≤ ≤ ≤ ≤ Assume q = 2, 9 n 20. If the involutions in G are Baer collineations ≤ ≤ PSL PSL of Π, then n = 16 by [22]. Since (3, 2) ∼= (2, 7) it follows from [12, 4.6(a), (b), 4.8(i)] that this cannot be the case. Hence the involutions in G are elations of Π, since they induce elations on = PG(2, 2). Therefore n is O ∼ even. Then n = 10, 12, 14, 16, 18, 20. The cases n = 10, 14 or 18 are ruled out by [24, Theorem 13.18]. The case n = 12 is ruled out by [27]. Hence either n = 16 or n = 20. Assume the latter occurs. Let be the set of points of Π E which do not lie on any secant to . Then = 288. Furthermore, G leaves O |E| E invariant. In particular the stabilizer in G of any point of has odd order, since E each involution has center in and axis a secant to . So, for any point P O O PSL ∈ E the group GP is isomorphic either to Z7 or to Z3 or to Z7.Z3, as G ∼= (2, 7). Thus the corresponding P G, which clearly lies in , has length 24, 56 or 8, E respectively. Thus 288 = = x 24 + x 56 + x 8. Now, let γ be a subgroup of |E| 1 2 3 h i G of order 7. As n = 20, then either γ fixes exactly 1 point on Π or a subplane h i of Π of order at least 6. Actually the latter cannot occur by [24, Theorem 3.7]. Hence γ fixes exactly one point of Π. Let Q be such a point. Clearly Q , since ∈ E = 288 and o(γ) = 7. On the other hand, by [41, Relation (9)], the group |E| γ fixes 3, 0 or 1 points on P G according to whether P G has length 24, 56 or h i 8, respectively. This in conjunction with the fact that γ fixes exactly one point h i in yields (x x , x ) = (0, 5, 1) in the above Diophantine equation. Let ρ be E 1, 2 3 h i a subgroup of G of order 3. Then ρ fixes exactly 2 points in each G-orbit of ACADEMIA h i PRESS length 56 or 8 by [41, Relation (9)]. Thus ρ fixes exactly 12 points on , as h i E x = 5 and x = 1. This yields that ρ fixes exactly a subplane of Π of order at 2 3 h i I I G

least 3, since n 2 mod 3. We have o(Fix( ρ )) 3, 4 by [24, Theorem 3.7]. ◭◭ ◮◮ ≡ h i ∈ { } Moreover, as any involution in G normalizing ρ is an elation of Π, the group G h i must induce an elation on Fix( ρ ). Thus the case o(Fix( ρ )) = 3 is ruled out. ◭ ◮ h i h i Hence o(Fix( ρ )) = 4. Now, let l be any line of Fix( ρ ). Then 3 (n 4), as h i h i | − ρ must be semiregular on l Fix( ρ ). A contradiction, since n = 20. Hence page 13 / 37 h i − h i n = 16 and part 4. of our statement follows from [12].

go back Assume q = 3, 28 n 34. Thus n cannot be a square. Then the involu- ≤ ≤ tions in G are homologies of Π, since they induce homologies of = PG(2, 3). O ∼ full screen Therefore n is odd and hence n = 29, 31 or 33. Actually the case n = 33 cannot occur by [24, Theorem 3.6]. Let be the set of points of Π which do not lie F close on any secant to . Then = (n 3)(n 9). Furthermore, G leaves in- O |F| − − F variant. In particular the stabilizer in G of any point of has odd order, since F quit each involution in G has center in and axis a secant to . Thus each G-orbit O O in must have length divisible by 16 as 16 G . Therefore 16 . That is F | | | | |E| 16 (n 3)(n 9). A contradiction since n = 29 or 31. This completes the | − − proof. PSU Lemma 3.4. Let Π be a ﬁnite projective plane of order n and let G ∼= (3, q) be a collineation group of Π with a point-orbit = (q), q > 2. If n ( ), O ∼ H ≤ O then G contains involutory perspectivities of Π. P

Proof. Let σ be an involution in G and suppose that it is a Baer collineation of Π. Assume that q is odd. Then σ fixes exactly q+1 points of lying on a secant l. O Furthermore, σ fixes a set S of q2 q secants and S l is a partition of the σ − σ ∪ { } points of (see [23]). The group C = C (σ) induces a transitive permutation O G group C¯ = PGL(2, q) on S , and C¯ = Z for any r S . Assume there exist ∼ σ r ∼ q+1 ∈ σ two distinct lines r and s of Sσ that intersect Fix(σ) (l ) in one and the same ∩ −O ¯ ¯ PGL point X. Then all the lines of Sσ pass through X, since Cr, Cs ∼= (2, q). Thus o(Fix(σ)) q2 q, since σ fixes S l . ≥ − σ ∪ { } Suppose that o(Fix(σ)) > q2 q. Then > 0, where denotes the set of − |E| E external lines to lying in Fix(σ) [X]. Clearly C acts on with kernel σ . In O ∩ E h i particular there exists a subgroup C of C such that C¯ = C / σ = PSL(2, q). 1 1 1 h i ∼ Clearly C¯1 acts in its natural 2-transitive permutation representation of degree q+1 on l . Furthermore, either C¯ fixes elementwise or C¯ has a non-trivial ∩O 1 E 1 orbit on . E Suppose that C¯ fixes elementwise. Let ρ¯ be any involution in C¯ . Then ρ¯ 1 E 1 fixes either 0 or 2 secants to through X according to whether q 1 mod 4 ACADEMIA O ≡ or q 3 mod 4 respectively, since C¯1 is transitive on Sσ and C¯1,u = Z q+1 for PRESS ≡ ∼ 2 each u S by [23]. ∈ σ I I G

Assume that q 1 mod 4. Thus ρ¯ is a Baer collineation on Fix(σ), since ◭◭ ◮◮ ≡ ρ¯ fixes exactly 2 points lying in l Fix(σ) and at least the point X ∩ O ⊂ ∈ Fix(σ) (l ). Furthermore, ρ¯ does not fix any secants to through X, other ◭ ◮ ∩ −O O than l. Thus √4 n + 1 = + 1, since ρ¯ fixes elementwise. Since √n + 1 = |E| E + q2 q + 1, we obtain √n = √4 n + q2 q. Thus √n = q2. A contradiction page 14 / 37 |E| − − by [41, Theorem 1.2], since l Fix(C¯ ) and > 0. E ∪ { } ⊂ 1 |E| go back Assume that q 3 mod 4. Then ρ¯ is a Baer collineation on Fix(σ), since ρ¯ ≡ fixes exactly 2 secants to through X other than l. Thus √4 n+1 = +3, since O |E| full screen ρ¯ fixes elementwise. Since √n + 1 = + q2 q + 1, we obtain √n √4 n = E |E| − − q2 q 2. That is − − close √4 n(√4 n 1)=(q + 1)(q 2) . (9) − − Clearly √4 n = q. Thus, either √4 n(√4 n 1) q(q + 1) or √4 n(√4 n 1) quit 6 − ≥ − ≤ (q 1)(q 2) according to whether √4 n q + 1 or √4 n q 1 respectively. A − − ≥ ≤ − contradiction in any case by (9). Suppose that C¯ has a non-trivial orbit on . Since the length of each C¯ -orbit 1 E 1 is the multiple of some non-trivial primitive permutation representation degree of C¯ , we have that d (C¯ ), where d (C¯ ) denotes the minimal of such 1 |E| ≥ 0 1 0 1 degrees of C¯ . If q / 3, 5, 7, 9, 11 , then d (C¯ ) = q + 1. Hence σ fixes at least 1 ∈ { } 0 1 q2 q+1+(q+1) lines through X, since √n+1 = +q2 q+1. So o(Fix(σ)) − |E| − ≥ q2 +1. That is √n q2 +1. On the other hand n q4 +2q2 +q +2 as n ( ) ≥ ≤ ≤ O and ( ) = q4 + 2q2 + q + 2 by Lemma 2.4(2). That is n < (q2 + 2)2 andP hence O √n

2. A contradiction by [24, Theorem 13.18], ≡ since C¯ = PSL(2, q) acts faithfully on Fix(σ). Thus q 3, 5, 7, 9, 11 . If q = 9, 1 ∼ ∈ { } 6 then o(Fix(σ)) q2 by the same argument as above, since d (C¯ ) = q. The ≥ 0 1 previous argument rules out the case o(Fix(σ)) q2 + 1. Hence o(Fix(σ)) = q2. ≥ That contradicts [41, Theorem 1.2], since C¯ fixes the flag (X, l) and q = 9. 1 6 Hence q = 9. Then √n 82, since n ( ) and ( ) = 6734. Note that ≤ ≤ O O √n = 82 cannot occur by the above argumentP involvingP [24, Theorem 13.18]. Hence √n 81. On the other hand, √n = + 72. Hence < 10. Recall that ≤ |E| |E| contains a non-trivial C¯ -orbit on . Then either = 6λ + µ for some λ 1 E 1 E |E| ≥ and µ 0, since the unique primitive permutation degree of PSL(2, 9) less than ≥ 10 is 6. Clearly λ = 1 as < 10. Note that µ denotes the number of lines of |E| fixed by the whole group C¯ . Moreover, each involution in C¯ fixes exactly 2 E 1 1 lines in the C¯ -orbit of length 6 lying in . Also, each involution in C¯ fixes the 1 E 1 line l. So the involutions in C¯1 are Baer collineations of Fix(σ). Thus √n = 81, since √n must be a square and 72 < √n 81. Then µ = 2. Hence consists ≤ E ACADEMIA of three C¯1-orbits of length 6, 1, 1. So Fix(C¯1) consists of more than one flag. PRESS Again a contradiction by [41, Theorem 1.2]. Now, assume that o(Fix(σ)) = q2 q. Let β¯ be any involution of C¯ . If q 3 − 1 ≡ I I G

mod 4, then β¯ fixes exactly two secants to through X, other than l. Thus β¯ is ◭◭ ◮◮ O a Baer collineation on Fix(σ) with o(Fix(β¯))=2. So q2 q = 4. A contradiction. − Hence q 1 mod 4 as q is odd. Then β¯ does not fix any secant to through X, ◭ ◮ ≡ O other than l. On the other hand, the collineation β¯ fixes exactly 2 points lying in l Fix(σ) and at least the point X Fix(σ) (l ). Then β¯ fixes a page 15 / 37 ∩ O ⊂ ∈ ∩ −O Baer involution of Fix(σ). A contradiction, since β¯ does not fix any secant to O go back through X other than l. Hence, we may assume that any two distinct lines of Sσ intersect l in distinct full screen points. Then l Fix(σ) contains l and the intersection points of l with ∩ ∩ O each line of S . Thus √n q2. Actually either √n = q2 or √n = q2 + 1, σ ≥ close since √n ( ) where ( ) = q4 + 2q2 + q + 2 by Lemma 2.4(2). If 2 ≤ O O √n = q , thenpPC¯1 acts transitivelyP on l Fix(σ) , since C¯1 is transitive on ∩ −O quit S . Hence l Fix(σ) consists of two C¯ -orbits of points of length q + 1 and σ ∩ 1 q2 q. A contradiction by [41, Theorem 1.2], since q is odd. If √n = q2 + 1, the − above argument involving Theorem 13.18 of [24] rules out this case. Thus the involutions in G are homologies of Π when q is odd. Assume that q is even. Let Q be a Sylow 2-subgroup of G containing σ. Then Z(Q) is an elementary abelian 2-group of order q fixing exactly a point Y in O and all the q2 secants to through Y (see [23]). Note that the involutions in O Z(Q) are Baer, since σ is a Baer involution and G contains a unique conjugate class of involutions. Thus n (q2 1)2 by [24, Theorem 3.7]. On the other ≥ − hand, we have that n ( ) = q4 + 2q2 + q + 2 by Lemma 2.4(2), and ≤ O hence n < (q2 + 2)2. So eitherP n = (q2 1)2 or n = q4 or n = (q2 + 1)2, as − n (q2 1)2. Thus the involutions in Z(Q) fix the same √n+1 lines through Y , ≥ − since they fixes the same q2 secants to through Y . Then Z(Q) is semiregular O on [Y ] Fix(Z(Q)). Therefore Q is semiregular on [Y ] Fix(Z(Q)), since each − − involution in Q lies in Z(Q) being q even. Thus q3 (n √n), being Q = q3 | − | | and [Y ] Fix(Z(Q)) = n √n. Then either q3 √n or q3 (√n 1). A | − | − | | − contradiction in any case, since √n = q2 1 or √n = q2 or √n = q2 + 1. Thus σ − is a perspectivity of Π. As a consequence each involution in G is a perspectivity PSU of Π, since G ∼= (3, q) has a unique conjugate class of involutions. This completes the proof. PSU Lemma 3.5. Let Π be a finite projective plane of order n and let G ∼= (3, q) be a collineation group of Π with a point-orbit ∼= (q), q > 2. If n ( ), O H PG ≤ 2 O then G acts strongly irreducibly on a G-invariant subplane Π0 ∼= (2, q P) of Π containing . In particular, the following hold: O 4 (1) Π = Π0 or n q ; ACADEMIA ≥ PRESS (2) the involutions in G are either homologies or elations of Π according to whether q is odd or even, respectively. I I G

Proof. If n q3 + 1, the assertion follows by [4, Theorem 3.10]. Hence we may ◭◭ ◮◮ ≤ assume that n > q3 + 1. We proceed stepwise.

◭ ◮ (I) G does not ﬁx any point of Π.

page 16 / 37 Suppose that G = G for some point P Π. Clearly P / . Then [P ] P ∈ ∈ O contains q3 + 1 tangents to , since G is primitive on . Then P cannot be go back O O the center of any perspectivity lying in G, since G is faithful on . Hence the O full screen axis of any involutory perspectivity contains P . Then q is even by [23], since any involution in G fixes q + 1 collinear points in for q odd. Furthermore, O close each point in is the center of an involutory perspectivity by [23]. Hence each O involution fixes at most 2 lines through P . quit Suppose that n is even. Then each line of [P ] tangent to is the axis of O some the involutory elation in G, since q is even. As n > q3 + 1, there exists an external line e to through P . Clearly G is odd. Then n + 1 q3 + 1 + eG , O | e| ≥ since [P ] = n + 1, eG [P ] and the group G is transitive on the tangents to | | ⊂ O2 through P . Moreover q3 eG , since q is even and G is odd. If G 3(q+1) , | | e| | e| ≤ j where j = (3, q + 1), then eG q3(q 1)(q2 q + 1)/3. A contradiction, since ≥ − 2 − eG [P ] and n q4. Hence G > 3(q+1) . Then q = 2 and G 9 by [16], ⊂ ≤ | e| j | e| | since G is odd. A contradiction, since q > 2 by our assumption. | e| Now, suppose that n is odd. For each Sylow 2-subgroup Q of G, with 1 i ≤ i q3 + 1, denote by U the center of Q . Then U = U (X ,z ) for some point ≤ i i i i i i X and some line z [P ], for each 1 i q3 + 1, since n is odd, q is i ∈ O i ∈ ≤ ≤ even and G is transitive on . Clearly X = X for each 1 w, j q3 + 1 O w 6 f ≤ ≤ with w = f. If there exist z = z for some 1 h, t q3 + 1 with h = t, 6 h t ≤ ≤ 6 then Ui is a Frobenius complement by [24, Theorem 4.25]. A contradiction, by [44, Proposition 18.1(i)], since q > 2. Hence there are exactly q3 + 1 external lines z to through P such that U = U (X ,z ). Now it easily seen that, the i O i i i i stabilizer of any external line, which is not a z , 1 i q3 + 1, must have odd i ≤ ≤ order. So, by using similar arguments to that used above, involving [16], we may conclude that the lines z , with 1 i q3 + 1, are the unique lines of [P ] i ≤ ≤ which are external to . Hence n +1=2(q3 + 1). That is n = 2q3 + 1. Hence O [P ] is the disjoint union of two 2-transitive G-orbits both of length q3 + 1. Let 2 Z = Z q+1 be a subgroup of G = PSU(3, q ), where j is defined as above. Clearly ∼ j ∼ Z fixes exactly 2(q + 1) lines of [P ]. Furthermore, Z fixes two points on each of these line other than P , since (2q3 + 1, q + 1) = 1 and (2q3, q + 1) = 1. Hence Z is planar. In particular Z fixes exactly q + 1 points of lying on a secant s. O ACADEMIA Furthermore, Z leaves a partition of q2 q + 1 secants invariant, with s , PRESS J − PGL ∈ J by [23]. In particular NG(Z) acts on Fix(Z), inducing NG(Z)/Z ∼= (2, q). Arguing as in Lemma 3.4, either there exists a point R Fix(Z) s such that ∈ ∩ I I G

[R], or any two distinct lines in intersect s in distinct points. This ◭◭ ◮◮ J ⊂ J −O yields o(Fix(Z)) q2 q in any case. A contradiction by [24, Theorem 3.7], ≥ − since n = 2q3 + 1, q is even and q > 2. ◭ ◮

page 17 / 37 (II) G does not ﬁx any line of Π.

Suppose that G = G for some line l of Π. Clearly l = . If l is the axis go back l ∩O ∅ of some involutory perspectivity in G, then q is even and hence n odd by [23]. Moreover G(l, l) = 1 by [24, Theorem 4.25], since G fixes l and acts on full screen 6 h i O transitively. Note that G(l, l) ⊳ G, since G fixes l. Then G = G(l, l), since G is close simple as q > 2 and G(l, l) = 1 . So G fixes l pointwise. A contradiction by (I). 6 h i Hence l contains the centers of all involutory perspectivities of G. Thus q is odd quit by [23]. Pick two distinct involutory perspectivities σ and φ in G with axis a and c respectively, such that a = c and such that S S share a secant m. 6 σ ∩ φ The secants in Sσ and in Sφ meet l in the center of σ and φ, respectively. In particular, both centers coincide with l m, since m S S . So σ fixes ∩ ∈ σ ∩ φ S S . A contradiction. σ ∪ φ (III) G acts irreducibly on Π.

Assume that G leaves invariant a triangle ∆. Then G ﬁxes ∆ pointwise, since G is simple as q > 2. A contradiction, since G does not ﬁx points by (I). Thus G is irreducible on Π.

PG 2 (IV) G acts strongly irreducibly on a G-invariant subplane Π0 ∼= (2, q ) of Π containing . In particular, either Π = Π or n q4. O 0 ≥ By Lemma 3.4 and by (III), we have that G is irreducible on Π and it contains perspectivities of Π. Then by [18, Lemma 5.3] the centers and the axes of the perspectivities in G generate a subplane Π containing = (q) and on which 0 O ∼ H G acts strongly irreducibly. Then o(Π ) q2 since Π contains all the q4 +q2 q3 0 ≥ 0 − secants of . Note that, again by [18], any other subplane of Π left invariant O by G contains Π0. Assume that q is odd. Then the involutions in G are homologies of Π by PG 2 Lemma 3.4, since Π0 ∼= (2, q ). Therefore, by [29, Theorem C(iii)], the group G leaves invariant a plane Π of order q2 containing a unital (q). In 1 H1 particular Π = PG(2, q2) and G acts on (q) in its natural action by [23]. 1 ∼ H1 Then Π = Π and (q) = (q), since Π Π and o(Π ) q2 by the above 0 1 H H1 0 ⊂ 1 0 ≥ ACADEMIA remark. At this point the assertion for q odd follows by [24, Theorem 3.7]. PRESS Assume that q is even. Then the involutions in G are elations of Π by [20, 2h 6h Proposition 6.1]. Suppose that o(Π0) > 2 . Then it must be o(Π0) > 2 by I I G

[7, Lemma 5.4]. A contradiction, since n ( ) = 24h + 22h+1 + 2h + 2 by ◭◭ ◮◮ PG 2h ≤ O Lemma 2.4 (2). Thus Π0 ∼= (2, 2 ) and againP the assertion follows by [24, Theorem 3.7]. ◭ ◮ PSU Lemma 3.6. Let Π be a ﬁnite projective plane of order n and let G ∼= (3, q), q page 18 / 37 odd, be a collineation group of Π with a point-orbit = (q). If n ( ), then O ∼ H ≤ O G contains involutory perspectivities and G is strongly irreducible on a PG-invariant go back subplane Π of Π containing . In particular, one of the following occurs: 0 O 2 PG 2 full screen (1) n = q , Π0 = Π ∼= (2, q ); 4 PG 2 (2) n = q , Π0 ∼= (2, q ) is a Baer subplane of Π. close Proof. If n q4, the assertion follows from Lemma 3.5. Hence assume that ≤ quit n > q4. We rule out this case in two steps.

(I) n = q4 + 2q2 2q. − Note that n q4 + q2 by [24, Theorem 3.7], since Π = PG(2, q2) and ≥ 0 ∼ n > q4. As q is odd the involutions in G are homologies of Π by Lemma 3.5(2). Then (q 1) (n 1) by [29, Theorem C(iii)]. Thus n = q4 + q2 + λ with − | − λ 1 and (q 1) (λ + 1). Then λ = λ (q 1) 1, λ 1, and hence ≥ − | 1 − − 1 ≥ n = q4 + q2 + λ (q 1) 1. Let Z(U) be the center of a Sylow p-subgroup 1 − − U of G. Then Z(U) is an elementary abelian p-group of order q. Furthermore Z(U) induces on Π an elation group having the same center P and the 0 ∈ O same axis c tangent to in P (see [23]). Let s be a secant to through P . O O Clearly s is a secant to Π and Z(U) is semiregular on s Π P . Assume that 0 ∩ 0 −{ } Z(U) = 1 for some point R on s Π . Then Z(U) is planar in Π, since U R 6 h i − 0 R is transitive on [P ] and Z(U) ⊳ U. Then each element in Z(U) is planar, ∩O R since they are conjugate in G. Let τ Z(U), τ = 1. Then o(Fix( τ )) q2, since ∈ 6 h i ≥ Fix(τ) c q2 + 1. Actually, either o(Fix( τ )) = q2 or o(Fix( τ )) = q2 + 1, | ∩ | ≥ h i h i since n ( ) = q4 + 2q2 + q + 2 by Lemma 2.4(2). ≤ O AssumeP that o(Fix( τ )) = q2 + 1. Note that the collineation ζ in G rep- h i resented by the matrix diag( 1, 1, 1) centralizes Z(U). Hence ζ and acts on − − Fix( τ ). In particular ζ acts non-trivially on Fix( τ ), since it acts non-trivially h i h i on Fix( τ ) by [23]. This is impossible by [24, Theorem 13.18], since h i ∩ O o(Fix( τ )) = q2 + 1 with q2 + 1 2 mod 4 and q2 + 1 > 2 as q is odd. h i ≡ Hence o(Fix( τ )) = q2. Then Fix(τ) c = q2 +1 and hence Fix(Z(U)) c = h i | ∩ | | ∩ | q2 + 1, since the non-trivial elements in Z(U) are conjugate in G and since Fix(Z(U)) c q2 + 1. Thus Z(U) is semiregular on c Fix(Z(U)). Hence | ∩ | ≥ − ACADEMIA q (n q2) as U = q and c Fix(Z(U)) = n q2. That is q n. Then PRESS | − | | | − | − | q (λ + 1), since n = q4 + q2 + λ (q 1) 1. Thus λ = λ q 1, λ 1, and | 1 1 − − 1 2 − 2 ≥ I I G

hence ◭◭ ◮◮ n = q4 +(λ + 1)q2 (λ + 1)q . (10) 2 − 2 ◭ ◮ Since n is odd, n ( ) = q4 + 2q2 + q + 2 by Lemma 2.4(2), it follows that 4 2 ≤ O 2 n q + 2q + q + 1P. Hence (λ2 1)q (λ2 + 2)q 1 0 by (10). This yields ≤ − − −4 ≤ 2 page 19 / 37 either λ2 = 1, or λ2 = 2 and q = 3. Thus either n = q + 2q 2q or n = 99 and − PSU q = 3. Assume the latter occurs. Let S be a Sylow 2-subgroup of G ∼= (3, 3). go back Then S = 25 and hence S ∤ 4(n + 1) as n = 99. A contradiction by [17, | | | | Satz 2], since n 3 mod 4. Thus the assertion (I). full screen ≡ (II) The final contradiction. close Let be the set of lines of Π which are external to Π . Denote by and E 0 S T quit the sets of the secants and of tangents to Π . Then and have size q4 +q2 +1 0 S T and (q4 + q2 + 1)(n q2) respectively. Now it is a straightforward computation − 2 to see that = 2q2 (q 1) q2 + q + 2 , since = n2 + n + 1 ( + ) |E| − |E| − |S| |T | and n = q4 + 2q2 2q. − Since the involutions in G are perspectivities of Π and of Π0, then Ge has odd order for each line e . Now assume there exists a line x in such ∈ E E that ( G , q+1 ) = 1, j = (q + 1, 3), and let γ be an element of order a prime | x| j divisor of q+1 . Then γ induces an (A, l)-homology group on Π by [23]. In j h i 0 particular Fix(γ) Π l = q2 + 1. Nevertheless γ is planar on Π, since | ∩ 0 ∩ | h i γ fixes x in . Note that x intersects l Π again since x . Thus γ h i E − 0 ∈ E h i fixes at least q2 + 2 points on l and hence o(Fix( γ )) q2 + 1 as γ is planar h i ≥ on Π. Arguing as above, with γ in the role of τ , we have o(Fix( γ )) = h i h i h i q2 + 1. Now, let σ be an involutions centralizing γ (we can pick σ in the h i same cyclic subgroup of G of order q+1 containing γ ). Then σ acts non- j h i trivially on Fix( γ ), since G has odd order and x Fix( γ ) . At this point h i x ∈ h i ∩ E the above argument involving [24, Theorem 13.18] rules out this case. Hence 2 G , q+1 = 1, j = (q + 1, 3), for each line e . That is (q+1) eG for | e| j ∈ E j 2 each line e (see [38]). Therefore (q+1) 2q2 (q 1)2 q2 + q + 2 , since ∈ E j | − consists of non-trivial G-orbits and = 2q2 (q 1)2 q2 + q + 2 . Actually E 2 |E| − (q+1) 16, as ( q+1 , q 1)=( q+1 , q2 + q + 2) = 2. Hence(q + 1)2 = 2ij, where j | j − j 1 i 4 and j =(q + 1, 3). Thus j = 1 and hence q 1 mod 3. At this point, ≤ ≤ ≡ easy computations show that no value of q is admissible. This completes the proof. PSU Lemma 3.7. Let Π be a finite projective plane of order n and let G ∼= (3, q) be ACADEMIA a collineation group of Π with a point-orbit = (q), q > 2. If n ( ), then PRESS O ∼ H ≤ O G contains involutory perspectivities and G is strongly irreducible on a PG-invariant subplane Π of Π containing . In particular, one of the following occurs: 0 O I I G

2 2 (1) n = q , Π0 = Π = PG(2, q ); ◭◭ ◮◮ ∼ 4 PG 2 (2) n = q , Π0 ∼= (2, q ) is a Baer subplane of Π. ◭ ◮ Proof. In order to obtain the assertion we have to investigate only the case q = 2h page 20 / 37 2 , since the assertion is true for q odd by Lemma 3.6. Hence, assume that q = 22h. Recall that the involutions in G are elations of Π by [20, Proposition 6.1]. go back Assume n 24h + 22h. Recall that n ( ) = 24h + 22h+1 + 2h + 2. If ≥ ≤ O n = 24h + 22h+1 + 2h + 2, then n 2 mod 4Pas h > 1. A contradiction by [24, ≡ full screen Theorem 13.18]. Then n 24h + 22h+1 + 2h, since n is even. We prove that this ≤ leads to a contradiction in ﬁve steps. close h (I) Let C = Z h , j = (2 + 1, 3). If there exists δ C, δ = 1, which is ∼ (2 +1)/j ∈ 6 quit planar on Π, then o(Fix(δ))=22h and hence Fix(δ) Π c = 22h + 1. | ∩ 0 ∩ | h PSL h Let C ∼= Z(2h+1)/j , where j = (2 + 1, 3). Then NG(C)/C = (2, 2 ) by PSL h [23]. Clearly NG(C) is the minimal such that NG(C)/C ∼= (2, 2 ). Since C is abelian and C ⊳ N (C), then C E C (C) E N (C). As N (C)/C G NG(C) G G ∼= PSL h (2, 2 ), then either CNG(C)(C) = C or CNG(C)(C) = NG(C). If the former

occurs, then NG(C)/C = NG(C)/CNG(C)(C) Aut(C). A contradiction, since PSL h ≤ C is cyclic while NG(C)/C = (2, 2 ). So CNG(C)(C) = NG(C). That is C Z(NG(C)). Furthermore NG(C)′ = NG(C) by the minimality of this one. ≤ h Hence NG(C) is perfect central extension of PSL(2, 2 ) by C. Then NG(C) = C L, where L = PSL(2, 2h) by [32, Theorem 5.1.4]. × ∼ Assume there exists δ C, δ = 1, which is planar on Π. Then o(Fix(δ)) 2h ∈ 6 PG 2h ≥ 2 , since C induces a group of homologies on Π0 ∼= (2, 2 ) having the same center X and the same axis l by [23]. Clearly L acts on Fix(δ), since L centralizes C. In particular L = PSL(2, 2h) has two orbits on Fix(δ) Π l ∼ ∩ 0 ∩ of length 22h + 1 and 22h 2h respectively (see [23]). If o(Fix(δ)) > 22h, then − o(Fix(δ))=22h + 1 by [24, Theorem 3.7], since n 24h + 22h+1 + 2h. Then L ≤ fixes the point D, where D = Fix(δ) l Π . Thus any Sylow 2-subgroup of L { } ∩ − 0 fixes exactly 2 points on Fix(δ) l, namely D and a point lying in the L-orbit of ∩ length 2h+1. Therefore any Sylow 2-subgroup of L, which is elementary abelian of order 2h, induces a group of homologies of Fix(δ). Note that L fixes X, since PG 2h L centralizes C and C induces a group of homologies on Π0 ∼= (2, 2 ) having the same center X and the same axis l. Hence L fixes the lines l and XD and the points X and D on Fix(δ). Thus distinct Sylow 2-subgroups of L are homology groups having either the same center and distinct axes, or distinct centers and the same axis. Indeed, distinct Sylow 2-subgroups of L fixes distinct points on ACADEMIA the L-orbit on l Π of length q + 1 (exactly one per each Sylow 2-subgroup PRESS ∩ 0 of L). Then the Sylow 2-subgroups of L are Frobenius complements by [24, Theorem 4.25]. Then h = 1 and q = 2, since they must be cyclic by [44, I I G

Proposition 18.1.(i)]. A contradiction, since q > 2 by our assumptions. Thus ◭◭ ◮◮ o(Fix(δ))=22h and hence assertion (I).

◭ ◮ (II) We have 4h 2h h+t n = 2 + 2 + 2 λ1 page 21 / 37 h t where 0 λ 2 − and t = (h + 1)/2 . ≤ 1 ≤ ⌊ ⌋ go back Let Z be the center of a Sylow 2-subgroup Q of G. Then Z is an elementary h full screen abelian group of order 2 and Z = Z(A, c) for some point A and some line ∈O h c of Π0 which is tangent to in A by Lemma 3.4 and by [23]. Thus 2 n. h O 3h h 4h 2h | close Then n = 2 λ for some positive integer λ 2 + 2 as n 2 + 2 . Thus ≥ ≥ λ = 23h + 2h + λ with λ 0. Hence n = 24h + 22h + 2hλ , λ 0. 0 0 ≥ 0 0 ≥ quit Assume λ 1. Since Z is semiregular on Π c, and since each involution 0 ≥ − in Q lies in Z as q = 2h, then also Q is semiregular on Π c. So 23h n2, since 3h 3h 2h 2 h 2 −4h 2h |h Q = 2 . Then 2 2 λ0 and hence 2 λ0, since n = 2 + 2 + 2 λ0. Thus | | t | | λ0 = 2 λ1, where λ1 1 and t = (h + 1)/2 (note that λ1 even number is also ≥ 4h 2h ⌊ h+t ⌋ 4h 2h+1 h admissible). Hence n = 2 +2 +2 λ1, λ1 1. As n 2 +2 +2 by the 4h 2h h≥+t ≤ h+t 2h h above argument, and being n = 2 + 2 + 2 λ1, we have 2 λ1 2 + 2 . t h h t ≤ That is 2 λ 2 + 1 and hence 1 λ 2 − , since λ is a positive integer 1 ≤ ≤ 1 ≤ 1 and t = (h + 1)/2 . ⌊ ⌋ 4h 2h Assume that λ0 = 0. Then it is easy to determine that n = 2 + 2 , which is assertion (II) for λ1 = 0.

(III) Let be the set of lines of Π which are external to Π . Then E 0 = 26h + 25h+tλ + 23h+tλ + 22h+2tλ2 , |E| 1 1 1 h t where 0 λ 2 − and t = (h + 1)/2 . ≤ 1 ≤ ⌊ ⌋ Let be the set of lines of Π which are external to Π . Denote by and E 0 S the sets of the secants and of tangents to Π , respectively. Then and T 0 S T have size 24h + 22h + 1 and (24h + 22h + 1)(n 22h), respectively. Therefore − + = (24h + 22h + 1)(n + 1 22h) and we obtain |S| |T | − + = (24h + 22h)n +(n + 1) 26h . (11) |S| |T | − As = n2+n+1 ( + ) and by bearing in mind that n = 24h+22h+2h+tλ , |E| − |S| |T | 1 λ 0, we have 1 ≥ 4h 2h h+t = (2 + 2 )n + 2 λ1n +(n + 1) ( + ) . (12) ACADEMIA |E| − |S| |T | PRESS Now, by combining (11) and (12), we have

= 2h+tλ n + 26h . (13) |E| 1 I I G

4h 2h h+t As n = 2 + 2 + 2 λ1, by (13) we obtain ◭◭ ◮◮ 6h 5h+t 3h+t 2h+2t 2 = 2 + 2 λ1 + 2 λ1 + 2 λ1 . (14) ◭ ◮ |E|

23h(2h+1)(22h 1) (IV) Each G-orbit in has length − , µ = 1 or 3. page 22 / 37 E µ

Observe that = y eG, y 1 and eG > 0, since G has odd order for go back E ∪i=1 i ≥ i ei each 1 i y. Hence ≤ ≤ y full screen = eG . (15) |E| i i=1 P close Set eG = min eG : 1 i y . Then y eG by (15). Hence i ≤ ≤ |E| ≥ quit y23h(23h + 1)(22 h 1) (16) Ge 6h 5h+t 3h+t − 2h+2t 2 , | | ≥ j(2 + 2 λ1 + 2 λ1 + 2 λ1)

by (II) and since G = 23h(23h + 1)(22h 1)/j, j = (2h + 1, 3). | | − h t As λ 2 − , we have 1 ≤ y23h(23h + 1)(22h 1) y23h(23h + 1)(22h 1) 6h 5h+t 3h+t − 2h+2t 2 6h+1 4h+1− j(2 + 2 λ1 + 2 λ1 + 2 λ1) ≥ j(2 + 2 ) and hence y(23h + 1)(22h 1) G − . (17) | e| ≥ 2h+1j(22h + 1)

h It is easily seen that Ge > max(9, 2 1) as h > 1. Now, assume that h | | − G , 2 +1 = 1. Then there exists a non-trivial element ψ in G of order | e| j 6 e 2h+1 2h+1 a divisor of j . Since the cyclic subgroups of G of order j are conjugate in G (indeed G is transitive on the lines on ), then we may assume that ψ C. O ∈ Then ψ is planar with o(Fix(ψ)) = 22h and Fix(ψ) Π c = 22h + 1 by (I). | ∩ 0 ∩ | Nevertheless ψ ﬁxes the external e to Π0, the secant c to Π0 and hence the 2 point e c. As e then e c lies in c Π0. Thus ψ ﬁxes at least q + 2 ∩ ∈ E ∩ − h points on C contradicting (I). Thus Ge is odd, Ge > max(9, 2 1) and 2h+1 | | | | 22h 2h+1− ( Ge , j ) = 1. Then Ge Z 22h−2h+1 .Z3 by [16] (note that ( −j , 3) = 1 | | ≤ j 22h 2h+1 by [18, Lemma 3.9]). Now, by using (17), it is plain to see that G > − | e| 3j 22h 2h+1 as h > 1. Hence −j Ge as Ge Z 22h−2h+1 .Z3. Actually, the previous | | | ≤ j 22h 2h+1 proof can be repeated for each G , 1 i y, in order to show that − ei ≤ ≤ j | ACADEMIA 22h 2h+1 and 2h− h . Hence , or , for each PRESS Gei Gei Z 2 2 +1 .Z3 Gei = µi −j µi = 1 3 | | ≤ j | | 3h h 2h G 2 (2 +1)(2 1) 1 i y. Therefore e = − , µi = 1 or 3, for each 1 i y. ≤ ≤ i µi ≤ ≤

I I G

(V) The ﬁnal contradiction. ◭◭ ◮◮

22h 1 G By (IV), we have − e for each 1 i y, since ◭ ◮ 3 | i ≤ ≤

G = 23h(2 3h + 1)(22h 1)/j, j = (2h + 1, 3) . page 23 / 37 | | − y 22h 1 G Then − as = e . Then go back 3 | |E| |E| i i=1 P full screen 22h 1 − 26h + 2tλ (25h + 23h + 22h+tλ ) 3 1 1 close 22h 1 t h+1 t 22h 1 by (III). Easy computations yield 3− (1+2 λ1(2 +2 λ1)). Hence f 3− = t h+1 t | h t quit 1 + 2 λ (2 + 2 λ ) with f 1, h > 1, t = (h + 1)/2 and 1 λ 2 − . A 1 1 ≥ ⌊ ⌋ ≤ 1 ≤ contradiction by Lemma 2.7.

We conclude that n = q4.

2G Lemma 3.8. Let Π be a ﬁnite projective plane of order n and let G ∼= 2 (q), q = 3h, h odd, h 1, be a collineation group of Π with a point-orbit = (q). If ≥ O ∼ R n ( ), then h = 1, G = PΓL(2, 8), = (3) and one of the following occurs: ≤ O ∼ O ∼ R (1) ΠP= PG(2, 8), G leaves a line oval of Π invariant and consists of the ∼ C O external points of ; C (2) n = 26.

Proof. Note that n ( ) = q4 + 2q2 + q + 2, where q = 3h and h is odd, ≤ O h 1 by Lemma 2.4(3).P Assume that n > q4. Then q4 < n < (q2 + 2)2, since ≥ ( ) < (q2 + 2)2. Let ψ be any involution of G = 2G (q). If ψ is a Baer O ∼ 2 Pinvolution of Π, then n is square. Then n =(q2 + 1)2, since q4 < n < (q2 + 2)2. Note that C (ψ) = ψ PSL(2, q) by [35]. Furthermore PSL(2, q) acts non- G h i × trivially on Fix(ψ), since Fix(ψ) contains a line s , s line of Π, and ∩ O ∩ O PSL(2, q) acts on s in its natural 2-transitive permutation representation of ∩O degree q + 1 (see [35]). So PSL(2, q) acts faithfully on Fix(ψ). A contradiction by [24, Theorem 13.18], since √n 2 mod 4 and √n > 2, being √n = q2 + 1 ≡ and q = 3h, h 1. Thus ψ is a perspectivity of Π. ≥ Assume h > 1. Then G = G(Q, l) by [20, Lemma 4.3] and G n. That is | | | q3(q3 + 1)(q 1) n. A contradiction, since n (q2 + 1)2 +(q + 1). − | ≤ PΓL Assume h = 1. Then G ∼= (2, 8). Let S be any Sylow 2-subgroup of G. If n is odd, then each involution in S is a homology of Π. Then S = S(P, l) ACADEMIA PRESS by [29, Lemma 3.1], since S ∼= E8, the elementary abelian group of order 8. A contradiction, since distinct involutions in G ﬁx distinct blocks of (3) pointwise R by [35]. Thus n is even and either S = S(X,X) for some point X Π (3), ∈ −R I I G

or S = S(a, a) for some external line a to (3), since S = E8. Actually, the ◭◭ ◮◮ R ∼ latter is ruled out by the above argument. Hence S = S(X,X). In particular the axis of each elation in S is a secant of (3) by [35]. Let U be another Sylow ◭ ◮ R 2-subgroup of G. Then U = U(Y, Y ) for some point Y of Π by the above −R1 argument with U in the role of S. Clearly G′ = S, U = PSL(2, 8). If X = Y , page 24 / 37 h i ∼ then G′ = G′(X,X) which is abelian by [24, Theorem 4.14], since distinct involutions in G have distinct axes which are secants of (3). A contradiction, go back R since G′ = PSL(2, 8). Hence X = Y . Then S and U fix the external line e = XY ∼ ′ ′ 6 G G full screen to 1. Hence G′ = S, U fixes e. Clearly X e and X = 9, since ′ R h i G ⊆ GX′ = E8.Z7. As n > 81, we have X ( e. Then GQ′ has odd order for each ∼ ′ ′ G G close Q e X , since each involution in G′ is an elation of center in X and axis a ′ ∈ − G secant to 1. Hence e X is union of non-trivial G′-orbits of length divisible 3 R 3 − quit by 2 , since 2 G′ . Then each of these orbits must have length divisible by | | | ′ either 56 or 72 by [1], since e XG 72. Consequently, since 81 < n < 104, − ≤ we have that n 8 = n + 1 9 must be divisible by 56 or by 72, a contradiction. − − Hence n q4 and the assertion follows from Theorem 2.6. ≤ PSL h Lemma 3.9. Let Π be a finite projective plane of order n and let G ∼= (2, 2 ), h 3, be a collineation group of Π with a point-orbit = (2h). If n ( ), ≥ O ∼ W ≤ O then P (1) Π = PG(2, 2h), the projective extension of is embedded in Π and the set ∼ O C of the external lines to is a line-hyperoval extending a line conic, or O (2) n = 22h, or (3) n = (2h + 1)2, h > 3, and the involutions in G are Baer collineations of Π.

Proof. If h = 3, then (3) = (8) by [47, Example 1.4], and hence assertions R ∼ W (1) and (2) follow from Lemma 3.8 in this case. Hence we may assume h > 3. Let S be a Sylow 2-subgroup of G. Then S is elementary abelian of order 2h ﬁxing a point P on Π, since n2 + n + 1 is odd. Actually P Π by [47, ∈ −O Lemma 7.1]. Assume that each involution in S is a Baer collineation of Π (recall that there exists a unique conjugate class of involutions in G). Then each involution in S ﬁxes exactly √n + 1 lines through P . Then

2h n +1+(2h 1)(√n + 1) | − by [24, Result 1.4]. Hence 2h (n √n). Thus either 2h (√n 1) or 2h √n. | − | − | Then either √n = 2h or √n = 2h + 1 or √n 2h+1. On the other hand, ACADEMIA h h 1 ≥ h+1 n ( ) = 2 (2 − 3+1) by Lemma 2.4(4). Thus the case √n 2 is ruled PRESS ≤ O ≥ out.P Hence either √n = 2h or √n = 2h + 1 and we have assertions (2) and (3), respectively. I I G

Now, assume that each involution in S is a perspectivity of Π. By [10], the ◭◭ ◮◮ group S fixes a pencil Φ of parallel lines in elementwise. Furthermore, for O each line s in Φ there exists a unique non-trivial element in S fixing s ◭ ◮ ∩ O pointwise. Hence Φ = 2h 1. Moreover S = S(Q, Q) for some Q on Π by | | − −O [29, Lemma 3.1], since S is abelian of order at least 8. Since n > 2h 1, there page 25 / 37 − exists a line a [Q] which is external to . Then S is semiregular on a Q , ∈ O − { } since the axis of each involution in S is a secant to . Thus 2h n. Now, let go back O | R be another Sylow 2-subgroup of G. Then R = R(C, C) by arguing as above with R in the role of S. Then G = R,S fixes the line QC, since both S and R full screen h i fix QC. Then the projective extension of is embedded in Π, since QG QC, O ⊆ close the point Q is the center of a pencil of parallel lines in and G is flag-transitive h h 1 h O 2h 1 on . Then either n +1=2 + 1, or n = 2 − (2 1), or n 2 − by [13, O h 1 h − ≥ h 1 h quit Proposition 2.1], since v = 2 − (2 1). Actually the case n = 2 − (2 1) is h − h 2h 1 h − ruled out, since 2 n. Hence either n = 2 or n 2 − and 2 n. | ≥ | h PSL h Assume that n = 2 . Then G ∼= (2, 2 ) is 2-transitive on QC, since G G h PG h Q = QC and Q = 2 + 1. Then Π ∼= (2, 2 ), by [36, Satz 3]. It is easily seen that there is exactly one external line to through each point of O QC, other than QC, since each Sylow 2-subgroup of G fixes a pencil of parallel lines in each of size 2h 1 by [10]. Let be the set of these external lines O − C to . Then = 2h + 1, since G is 2-transitive on . Thus either there exists O |C| C a point X Π QC such that [X], or is a line conic of Π. If [X], ∈ − C ⊂ C C ⊂ then G fixes X and hence S = S(Q, QX). A contradiction, since S consists of elations having the same center Q but distinct axes which are secants to . O Hence is a line conic and QC is a line-hyperoval of Π extending . That C C ∪ { } C is assertion (1). 2h 1 h G Assume that n 2 − and 2 n. Then Q ( QC. Let B be a point ≥ | on QC which is not the center of any pencil of parallel lines to . Then G O B must have odd order. Hence GB Z2h 1 by direct inspection of the list of ≤ ± G 2h 1 subgroups of G given in [25, Haupsatz 8.27]. Furthermore B 2 − 3, h h 1 G h 2h ≤h 1 since n 2 (2 − 3+1) and Q = 2 + 1. Thus GB (2 1) /2 − 3 and ≤ G| | ≥h h − hence GB = Z2h 1, as GB Z2 h 1. That is B = 2 (2 1). Let x and ∼ ± ≤ ± ∓ − x be the number of G-orbits on QC QG of length 2h(2h 1) and 2h(2h + 1) + − − respectively. Then

h h h h h 2 (2 1)x + 2 (2 + 1)x+ = n 2 , (18) − − − G h h h 2h 1 since QC Q = n 2 . In particular (x + x+)2 (2 1) 2 − 3 − h h−1 − h+1 −h ≤ h 1 by (18 ), since n 2 (2 − 3+1). If x + x+ 2, then 2 (2 1) 2 − 3. ≤ − ≥ − ≤ A contradiction. Thus x + x+ = 1 and hence either (x , x+) = (1, 0) or ACADEMIA − h h − h PRESS (x , x+) = (0, 1). Assume the latter occurs. Then 2 (2 + 1) = n 2 by (18) − − and hence n = 22h + 2h+1. So, QC consists of two G orbits of length 2h + 1 h h and 2 (2 + 1). Then the group GB = Z2h 1 ﬁxes exactly 4 points on QC, ∼ − I I G

namely two in each G-orbit on QC. Furthermore there exists a subgroup GB∗ of ◭◭ ◮◮ G , such that [G : G∗ ] 3, ﬁxing at least 4 lines, including QC, through B, B B B ≤ since (n, 2h 1) = 3 and being n = 22h + 2h+1. Let η be any involution in G ◭ ◮ − normalizing . Then normalizes , since h . In particular GB η GB∗ GB ∼= Z2 1 η moves as G h h . Thus is planar and − , since page 26 / 37 B B = 2 (2 + 1) GB∗ o(Fix(GB∗ )) = 3 G (and hence G ) ﬁxes exactly 4 points on QC. Clearly η acts non-trivially on B B∗ go back Fix(GB∗ ). Thus η induces a homology on Fix(GB∗ ) as o(Fix(GB∗ ))=3. A contra- 2h diction, since η is an elation of Π. Thus (x , x+) = (1, 0) and hence n = 2 . − full screen That is assertion (2). Corollary 3.10. If case (2) or (3) of Lemma 3.9 occurs and if the involutions in G close are Baer collineations of Π when case (2) occurs, then the following hold:

quit (1) Each Sylow 2-subgroup of G induces a group of perspectivities having the same center and the same axis on the Baer subplane ﬁxed by any one of its involution; (2) If n = (2h + 1)2 is a prime power, then n is actually the square of a Fermat prime.

Proof. Assume either n = 22h or n = (2h + 1)2, and that the involutions in G are Baer collineations of Π. We may also assume h > 3 by Lemmas 3.8 and 3.9. Let S be a Sylow 2-subgroup of G. Then S is elementary abelian of order 2h. As G = D h for each P , then Fix(α) = Fix(β) for α, β S 1 , with P ∼ 2(2 +1) ∈ O 6 ∈ − { } α = β. In particular Fix(S) Fix(σ) for each σ S 1 . Furthermore, since 6 ⊂ ∈ − { } b = 22h 1 and since G is transitive on the lines of , the group S fixes exactly − O 2h 1 lines of . − O Assume that β induces a Baer involution on Fix(α). Then √n must be a square. If √n = 2h + 1, then (√4 n 1)(√4 n + 1) = 2h. As (√4 n 1, √4 n + 1) = 2, 4 4 − h 1 − then √n 1 = 2 and √n +1=2 − . A contradiction since h > 3. Hence − √n = 2h and h is even. Note that S acts on the plane Fix(α) Fix(β) fixing at ∩ least √n 1 = 2h 1 lines on it. Clearly among these lines there are 2h 1 − − − secants to . As √n 1 > √4 n + √8 n + 1, and since the order of Fix(α) Fix(β) O − ∩ is √4 n (recall that Fix(α) = Fix(β)), we have Fix(S) = Fix(α) Fix(β). Let l be 6 ∩ any line of Fix(S). Then each non-trivial involution in S fixes exactly √n √4 n − points on l which are not fixed by any other involution in S. Then S must be semiregular on the non-empty point-set l σ S 1 (Fix(σ) l). Hence −∪ ∈ −{ } ∩ 2h (n + 1 (2h 1)(√n √4 n) (√4 n + 1)) . | − − − − h 4 +1 Hence 2h (n + √n 2√n). As √n = 2h and h is even, we have 2h 2 2 . This ACADEMIA | − | PRESS yields h = 2. A contradiction since h > 3. Thus each non-trivial element in S induces a perspectivity on Fix(S). Actually, as S fixes exactly 2h 1 > 3 lines − I I G

of and since S is a abelian, the group S induces the group S/ α on Fix(α) ◭◭ ◮◮ O h i consisting of perspectivities having the same center C and the same axis a. We have thus assertion (1). ◭ ◮ Now, assume that n = (2h + 1)2 is a prime power, then 2h + 1 = uj for page 27 / 37 some prime u and some j 1. Then j = 1 and u is a Fermat prime by [45, ≥ Result (B1.1)], as h > 3. We have thus assertion (2). go back Example 3.11. Let Π = PG(2, 2h) be a subplane of Π = PG(2, 22h) and let be 0 ∼ ∼ C a line hyperoval of consisting of a line-conic and an additional line . Clearly full screen Π0 0 l h C H = PSL(2, 2 ) fixes l and acts 2-transitively on 0. By [10], the external lines C ∼ C to and the points of Π define an incidence structure which is left invariant close C − C O by H . In particular = W (2h) and H acts flag-transitively on . C O ∼ C O quit The first part of Theorem 1.1 is now a consequence of the results in this section.

4. The non-faithful action

Throughout this section we assume that N = 1 . Then N is planar on Π, since 6 h i Fix(N) and is a non-trivial 2-(v, k, 1) design. We may also assume that O ⊆ O G is minimal with respect to the property that G/N is ﬂag transitive on . O Lemma 4.1. Then N = Φ(G), where Φ(G) is the Frattini subgroup of G.

Proof. Let S be a Sylow t-subgroup of N. Then S ⊳ G by the minimality of G, since G = NG(S)N by the Frattini’s argument. Thus N is nilpotent. Suppose that N Φ(G). Then there exists a maximal subgroup H of G such that G = 6≤ NH by [25, Satz 3.2(b)]. Clearly H is ﬂag-transitive on . Furthermore H < G O and H = G . A contradiction by the minimality of G. Hence N Φ(G). Note H N ∼ N ≤ that G∩ is maximal in G for each point P , since N ⊳ G and since G/N P ∈ O P is primitive on by [21]. Hence Φ(G) ⊳ G for each point P . Thus O P ∈ O N = Φ(G).

Lemma 4.2. If G/N is non-abelian simple, then one of the following holds: (1) G is a perfect central extension of G/N, or (2) there exists a Sylow t-subgroup S of N such that G/N SL(W ), where ≤ W = S/Φ(S) and Φ(S) is the Frattini subgroup of S.

ACADEMIA Proof. If N Z(G) we have the assertion (1), since G = G′ by the minimality PRESS ≤ of G. Hence we may assume that N Z(G). Then there exists a Sylow t-sub- 6≤ group S of N such that S Z(G), since N is nilpotent. Set W = S/Φ(S), where 6≤ I I G

Φ(S) is the Frattini subgroup of S. Clearly G acts on W . Let R be the kernel of ◭◭ ◮◮ the action of G on V . If U is the Sylow u-subgroup of N, where u is a prime, u = t, then [S, U] = 1 , since N is nilpotent. This yields N E R E G, since ◭ ◮ 6 h i S′ Φ(S), S being a t-group. If R = G, then each Sylow j-subgroup of G, with ≤ j = t, centralizes S by [15, Theorem 5.1.4]. That is C (S) N. Furthermore, page 28 / 37 6 G CG(S) ⊳ G as S ⊳ G. Then N ⊳ CG(S)N E G. Hence G = CG(S)N, since go back G/N is non-abelian simple and since CG(S) N. Actually, G = CG(S) since N = Φ(G) by Lemma 4.1. A contradiction, since S Z(G). Hence R < G. 6≤ Then R = N as G/N is non-abelian simple. Then G/N ΓL(W ), since W is full screen ≤ a vector space over GF(t). Actually G/N SL(W ), since G/N is non-abelian ≤ close simple. Thus we have assertion (2).

Corollary 4.3. In case (2) of Lemma 4.2, we have S 1 + d0(G/N), where quit | | ≥ d0(G/N) is the minimal primitive permutation representation of G/N.

Proof. By Lemma 4.2(2) there exists a Sylow t-subgroup S of N such that G/N SL(W ), where W = S/Φ(S) and Φ(S) is the Frattini subgroup of S. ≤ In particular G/N PSL(W ) as G/N is non-abelian simple. Then PG(W ) ≤ | | ≥ d0(G/N), where d0(G/N) denotes the minimal primitive permutation representation of G/N. This yields W 1+d (G/N) and hence S 1+d (G/N). | | ≥ 0 | | ≥ 0 It should be pointed out that the lower bound for S given in the previous | | corollary is not the best one. Indeed, stronger inequalities are given in Theo- rem 5.3.9 and Proposition 5.4.11 of [32]. Lemma 4.4. Let Π be a projective plane of order n and let G be a collineation group of Π with a point-orbit of length v. Assume that has the structure of O O a non-trivial 2-(v, k, 1) design, the group G induces a ﬂag-transitive and almost simple automorphism group on and n ( ). If G does not act faithfully on O ≤ O , then one of the following holds: P O (1) n q4, = (q), q > 2, Fix(N) = PG(2, q2) and G/N = PSU(3, q); ≥ O ∼ H ∼ ∼ (2) n q2, = (q), q = 22h, h 3, Fix(N) = PG(2, q) and G/N = ≥ O ∼ W ≥ ∼ ∼ PSL(2, q); (3) Π is the generalized Hughes plane over the exceptional nearﬁeld of order 72, = PG(2, 7) and SL(3, 7) G; O ∼ ≤ (4) n > q3, q = 2 or 3, = Fix(N) = PG(2, q) and G/N = PSL(3, q). O ∼ ∼ Proof. Assume that ( Fix(N). Clearly Fix(N) is a proper subplane of Π, O since N = 1 . Clearly G/N acts faithfully on Fix(N). If m is the order of ACADEMIA 6 h i PRESS Fix(N), then from [24, Theorem 3.7] and our assumption we obtain

m √n ( ) . (19) ≤ ≤ qX O I I G

If = PG(d, q), d 2, then q < m ( ) (note that the first inequality ◭◭ ◮◮ O ∼ ≥ ≤ O follows from the fact that ( Fix(N)). Now,pP we may use the same argument of O Lemma 3.2 involving transvections and Theorem 3.7 of [24] in order to obtain ◭ ◮ 2 qd 1 − 1 m ( ) . (20) page 29 / 37 q 1 − ≤ ≤ O − qX go back Using Lemma 2.4(1), we obtain (qd q)4 q2d+1 + qd+3 + qd+2 2qd+1 2qd + q4 4q2 + 4 full screen − − − − . (21) (q 1)4 ≤ (q2 1) (q 1) − − − close Now, arguing as in the first part of Lemma 3.2, we reduce to the case d = 2 or 3. These values are ruled out, since they do not satisfy (21) as (d, q) = (2, 2), (2, 3). 6 quit Assume either = (q), q > 2, or = (q), q = 32h+1. Then m O ∼ H O ∼ R ≤ q4 + 2q2 + q + 2 by (19) and by Lemma 2.4(2)–(3). This yields m < q2 + 2. Then,p by the first part of Theorem 1.1, with Fix(N) in the role of Π, we have = (q), q > 2, Fix(N) = PG(2, q2) and by G/N = PSU(3, q), since N = 1 O ∼ H ∼ ∼ 6 h i and m < q2 + 2. That is assertion (1). h Assume ∼= (q), q = 2 , h 3. Then m q(3q + 1)/2 by (19) and O W 2≥ ≤ PG Lemma 2.4(2). This yields m < q . Then m = q andp hence Fix(N) ∼= (2, q) by the first part of Theorem 1.1. Thus we obtain assertion (2).

Finally, assume = Fix(N). Then is symmetric and hence = Fix(N) ∼= PG O O PSL O (2, q) by Theorem 2.2. Therefore G/N ∼= (3, q), since G/N is ﬂag- transitive on and G is almost simple. If n q3, then = PG(2, 7), Π is O ≤ O ∼ the generalized Hughes plane over the exceptional nearﬁeld of order 72 and SL(3, 7) G by Theorem 2.5. That is assertion (3). If n > q3, then q = 2 or 3, ≤ since n ( ) = 2q2 +4q+4 by our assumption and Lemma 2.4(1). Therefore ≤ O we have assertionP (4). Lemma 4.5. The case n > q3, q = 2 or 3, = Fix(N) = PG(2, q) and G/N = O ∼ ∼ PSL(3, q) does not occur.

Proof. Assume = Fix(N) = PG(2, q), q = 2 or 3, G/N = PSL(3, q) and O ∼ ∼ n > q3. If q = 3, then 28 n 34, since q3 < n < ( ) = 2q2 + 4q + 4 by ≤ ≤ O Lemma 2.4(1). Thus n cannot be a square and hence NPhas odd order. Further- more, the involutions in G are homologies of Π, since they induce homologies on = PG(2, 3). At this point we may use the same argument of Lemma 3.3 to O ∼ rule out this case. Hence q = 2. Then 9 n 20. ≤ ≤ Assume there exists δ N, δ = 1, such that Fix(N) ( Fix(δ). Then either ∈ 6 ACADEMIA o(Fix(δ)) = 4 or o(Fix(δ)) 6 by [24, Theorem 3.7]. If the latter occurs, then PRESS ≥ n 36 again by [24, Theorem 3.7]. A contradiction, since n 20. Hence ≥ ≤ o(Fix(δ))=4, then n = 16 by [46], since n 20. ≤ I I G

Now consider N(δ) = α N : Fix(α) = Fix(δ) . Then N(δ) is a non-trivial ◭◭ ◮◮ { ∈ } subgroup of N, since Fix(δ) is a Baer subplane of Π. Clearly N(δ) = N(α) for each α N(δ). Thus we may set N0 = N(δ). Clearly Fix(N0) = Fix(δ) and ◭ ◮ ∈ hence Fix(N ) is a Baer subplane of Π. Assume there exists g G such that 0 ∈ g . Thus g , since and hence g page 30 / 37 N0 N0 = 1 Fix(N0) = Fix(N0 ) Fix(N0) Fix(N0 ) ∩ 6 h i g are Baer subplanes of Π. Then N0 = N0 . Hence distinct conjugates of N0 in G go back have trivial intersection and the corresponding Baer subplanes intersect exactly in Fix(N). full screen Now, consider N G and N N . Then N N N G , since N ⊳ G. In particular, 0 0 0 | 0 it is easy to see that

close N G G / N (N ) G/N 0 = | | | G 0 | = | | . quit N N N / N (N ) N/(N (N ) N) 0 | | | N 0 | | G 0 ∩ |

Since N/(N (N ) N ) = N (N )N/N, then G 0 ∩ ∼ G 0 G N N0 / N0 = [G/N : NG(N0)N/N] . (22)

G N If [G/N : NG(N0)N/N] = 1, then N0 = N0 by (22). Hence N0 ⊳ G. In particular N ⊳ N and hence N acts as N/N on Fix(N ). Since N/N must be 0 0 0 0 semiregular on u (Fix(N0) Fix(N0)), where u is any secant of Fix(N), we ∩ − PSL SL have [N : N0] = 2 and hence G/N0 ∼= Z2. (2, 7). That is G/N0 ∼= (2, 7). A contradiction by [1], since G/N PΓL(3, 4) being Fix(N ) = PG(2, 4). Thus 0 ≤ 0 ∼ [G/N : NG(N0)N/N] > 1 and hence [G/N : NG(N0)N/N] 7, since the mini- ≥PSL mal primitive permutation representation degree of G/N ∼= (2, 7) is 7. Then N G 7 N N by (22). Let l be a secant line of Fix(N). Note that, by the 0 ≥ 0 above argument different conjugates of N in G cover different pairs of points 0 on l Fix(N), since l (Fix(N ) Fix(N)) = 2. Hence 2 N G l Fix(N) . − | ∩ 0 − | 0 ≤ | − | That is 14 N N 14, since N G 7 N N and l Fix(N ) = 14. This yields 0 ≤ 0 ≥ 0 | − | N G = 7 , N N = 1 and hence N ⊳ N . Then [N : N ] = 2 arguing as 0 0 0 0 above. Suppose there exists a distinct conjugate N of N in G. By the above 0∗ 0 argument with N0∗ in the role of N0, we have [N : N0∗] = 2. This yields that N ∼= E4, since [N : N0] = 2 and N N0∗ = 1 . Thus N Z(G) and hence PSL∩ h i ≤ G is a perfect central extension of (2, 7) by Lemma 4.2. Then N ∼= Z2 by [32, Theorem 5.1.4], since N = 1 by our assumptions. Therefore N must 6 h i 0 be trivial. A contradiction, since δ N and δ = 1. Thus N ⊳ G and hence ∈ 0 6 0 N0 ⊳ N. Then the above argument rules out this case. Hence, we may assume that N is semiregular on l Fix(N), where l is the above secant. If N has even − order, then N contains Baer involutions of Π, since Fix(N) = = PG(2, 2). O ∼ ACADEMIA Then n = 4, since N is semiregular on l Fix(N). A contradiction, since PRESS − n > 23 by our assumption. Hence N has odd order. Thus each involution of G acts faithfully on . Then each such involution is either a Baer collineation O I I G

or an elation of Π, since = PG(2, 2). Thus either n is a square or n is even. ◭◭ ◮◮ O ∼ Then n = 9, 12, 14, 16, 18, 20, since 9 n 20. The cases n = 10, 14 or 18 ≤ ≤ are ruled out by [24, Theorem 13.18]. The case n = 12 is ruled out by [26]. ◭ ◮ Hence n = 9, 16 or 20. Then either N ∼= Z7 for n = 9 or 16, or N ∼= Z5 for n = 9, since N l Fix(N) again by the semiregularity of N, since page 31 / 37 | | | | − | l Fix(N) = n 2 and since N is non-trivial of odd order. Then N Z(G) | − | − PSL ≤ go back and hence G is a perfect central extension of (2, 7) by Lemma 4.2. Then PSL G = G0 N, where G0 ∼= (2, 7) by [32, Theorem 5.1.4], as N has odd × PSL full screen order. Therefore G = G0 ∼= (2, 7), since N = Φ(G) by Lemma 4.1. A contradiction, since N = 1 by our assumptions. 6 h i close Lemma 4.6. Let Π be a projective plane of order n and let G be a collineation group of Π with a point-orbit . Assume that has the structure of a non- quit O O trivial 2-(v, k, 1) design, the group G induces a flag-transitive and almost simple automorphism group on and n ( ). If G does not act faithfully on , then O ≤ O O one of the following holds: P (1) = PG(2, 7), Π is the generalized Hughes plane over the exceptional nearfield O ∼ of order 72 and SL(3, 7) G; ≤ (2) we have n q4, q 2 mod 3, q > 2, = (q), Fix(N) = PG(2, q2) and ≥ ≡ O ∼ H ∼ SU(3, q) G. ≤ Proof. By Lemmas 4.4 and 4.5, one of the following occurs: (1) = PG(2, 7), Π is the generalized Hughes plane over the exceptional O ∼ nearfield of order 72 and SL(3, 7) G; ≤ (2) = (q), q > 2, Fix(N) = PG(2, q2) is a Baer subplane of Π and G/N = O ∼ H ∼ ∼ PSU(3, q); (3) = (q), q = 2h, h 3, Fix(N) = PG(2, q) and G/N = PSL(2, q). O ∼ W ≥ ∼ ∼ As case (1) does indeed occur, we shall focus on the remaining two cases. Suppose that N Z(G). Then G is a perfect central extension of G/N by ≤ Lemma 4.2(1). Therefore G = SU(3, q), N = Z and q 2 mod 3 by [32, The- ∼ 3 ≡ orem 5.1.4], since N = 1 by our assumption. Thus we obtain assertion (2). 6 h i Suppose that N Z(G). Then, by Corollary 4.3, there exists a Sylow t-subgroup S of N such that S 1 + d (G/N), where d (G/N) is the minimal | | ≥ 0 0 primitive permutation representation of G/N. Assume that = (q), q > 2. As d (G/N) = q3 +1 for q = 5 and d (G/N) = O ∼ H 0 6 0 51 for q = 5 by [32], we have that S q3 + 2 for q = 5 and S 51 for q = 5, | | ≥ 6 | | ≥ ACADEMIA respectively. Moreover, it is easy to see that Fix(N) = Fix(α) for each non- PRESS trivial α N by [24, Theorem 3.7]. Thus S is semiregular on s s Fix(N) ∈ − ∩ I I G

for each secants s to . Therefore S q2(q2 1). A contradiction, since S is a ◭◭ ◮◮ O | | | − t-group with either S q3 + 2 for q = 5 or S 51 for q = 5. | | ≥ 6 | | ≥ h ◭ ◮ Assume that ∼= (2 ), h 3. Let α N, α = 1. Then α is planar, OPG Wh ≥ ∈ 6 2h since Fix(N) ∼= (2, 2 ). If ( Fix(α), then o(Fix(α)) 2 and hence 4h O ≥ h h 1 page 32 / 37 n 2 by [24, Theorem 3.7]. A contradiction since n ( ) = 2 (2 − 3+1) ≥ ≤ O by Lemma 2.4(4). Thus Fix(N) = Fix(β) for each βP N, β = 1. Then h ∈ 6 go back N (n 2 ), since N is semiregular on l (l ) for any secant line l to . | | | − h 2h−1 ∩O h h 1 O In particular S (n 2 ). Then S 2 − 3 as n 2 (2 − 3+1). If t = 2, | | | − | | ≤ ≤ 6 full screen instead of using the inequality S 1 + d0(G/N), we use the inequality S 2h 1 2h 1 | | ≥ | | ≥ 2 − , as S/Φ(S) 2 − by [32, Theorem 5.3.9], since G/N PSL(S/Φ(S)) | | ≥ 2h 1 2h 1 2h 1 ≤ 2h 1 close and h 3. Then S 2 − and hence 2 − 2 − 3, as S 2 − 3 ≥ | | ≥ ≤ | | ≤ by the above argument. It is a straightforward computation to show that the 2h 1 2h 1 quit inequality 2 − 2 − 3 is impossible, as h 3. So t = 2. Thus N contains ≤ 2h ≥ PG h Baer involutions. This yields n = 2 , since Fix(N) ∼= (2, 2 ) and since Fix(N) = Fix(β) for each β N, β = 1. Moreover N 2h(2h 1), since N is ∈ 6 | | | − semiregular on l (l ) for any secant line l to . In particular S 2h(2h 1). − ∩O O | | | − A contradiction, since S is a 2-group and since S 2h + 2 by Corollary 4.3, h | | ≥ being d0(G/N) = 2 + 1. This completes the proof. Theorem 4.7. Let Π be a projective plane of order n and let G be a collineation group of Π with a point-orbit of length v. Assume that has the structure of O O a non-trivial 2-(v, k, 1) design, the group G induces a ﬂag-transitive and almost simple automorphism group on and n ( ). If G does not act faithfully on O ≤ O , then one of the following holds: P O (1) Π is the generalized Hughes plane over the exceptional nearﬁeld of order 72, = PG(2, 7) and SL(3, 7) G; O ∼ ≤ (2) we have n = q4, q 2 mod 3, q > 2, = (q), Fix(N) = PG(2, q2) and ≡ O ∼ H ∼ SU(3, q) G. ≤ Furthermore, the involutions in G are perspectivities of Π.

Proof. As a consequence of Lemma 4.6, in order to complete the proof of this theorem, we need to investigate only the case n q4, q 2 mod 3, q > 2, ≥ ≡ = (q), Fix(N) = PG(2, q2) and SU(3, q) G. O ∼ H ∼ ≤ We proceed stepwise.

(I) The involutions in G are either homologies or elations of Π according to whether q is odd or even respectively. ACADEMIA PRESS SU Note that the group G ∼= (3, q) has a unique conjugate class of involutions, since N ∼= Z3 is central in G. Hence assume that the involutions in G are Baer I I G

collineations of Π. Then either n = q4 or (q2 + 1)2, since q4 n ( ) = ◭◭ ◮◮ ≤ ≤ O q4 + 2q2 + q + 2 by Lemma 2.4(2). P 2 2 ◭ ◮ Assume that q is odd. The case n =(q + 1) cannot occur by Theorem 13.18 of [24], since for any involution in γ in G the group CG(γ) induces a group con- 4 page 33 / 37 taining PGL(2, q) on Fix(γ). Thus n = q . Then the involutions in G are homologies of Π by [41, Proposition 2.6], and since G contains a unique conjugate class go back of involutions. A contradiction by our assumption. Therefore q is even. Now, let Z be the center of a Sylow 2-subgroup Q of G. Then Z is an elementary abelian PG 2 full screen 2-group inducing on Fix(N) ∼= (2, q ) an elation group having the same center A and the same axis c which is tangent to in A (see [23]). Thus ∈ O O close Fix(Z) Fix(N) c = q2 + 1. Then √n + 1 = Fix(Z) c = q2 + 1 or q2 + 2 | ∩ ∩ | | ∩ | according to whether n = q4 or (q2 + 1)2 respectively, since the involutions in quit G are Baer collineations of Π. Therefore Z is semiregular on c Fix(Z). Then − Q is semiregular on c Fix(Z), since each involution in Q lies in Z and q is − even. So q3 (n √n), since Q = q3 and c Fix(Z) . Thus either q3 √n or | − | | | − | | q3 (√n 1). A contradiction in any case, since √n = q2 or q2 + 1. Thus we | − PG 2 have assertion (I), since Fix(N) ∼= (2, q ).

(II) n = q4.

As n q4 we have either n = q4 or n q4 + q by [24, Theorem 3.7]. If the ≥ ≥ latter occurs, then the set of the external lines to Fix(N) is non-empty. Now, E it is a plain argument to see that the proofs of Lemmas 3.6 and 3.7 still work, with Fix(N) in the role of Π0, leading to a contradiction. Indeed, N ∼= Z3 is semiregular on and N is disjoint from any Sylow p-subgroup of G, p a prime E divisor of q, as q 2 mod 3. Thus n = q4 and hence the assertion. ≡ The proof of the second part of Theorem 1.1 is now complete.

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page 37 / 37 Alessandro Montinaro DIPARTIMENTO DI MATEMATICA, UNIVERSITA´ DEGLI STUDI DI LECCE, 73100 LECCE, ITALY go back e-mail: [email protected]

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